Problem and Solution – Limit at Infinity

       Here are the “all-out” problems on limit at infinity. Every problem is already attached by the solution, so don’t worry if you get stuck. For your information, the problems are collected from various mathematics literatures.

Read: Problem & Solution – Limit of Trigonometric Functions

The following formulas (theorems) are often applied to solve the problems related to this matter.

Limit at Infinity’s Formulas

The Infinity and The Zero
$\displaystyle \lim_{x \to \infty} \dfrac{1}{x^n} = 0$ for $n \geq 1$
Polynomial Function
If $f(x)$ and $g(x)$ be polynomial functions, then

$$\begin{aligned} \displaystyle \lim_{x \to \infty} \dfrac{f(x)}{g(x)} = \begin{cases} 0, &~\text{if the degree of}~f(x) < g(x) \\ \dfrac{\text{The coefficient of a leading term on}~f(x)}{\text{The coefficient of a leading term on}~g(x)}, &~\text{if the degree of}~f(x) = g(x) \\ \infty, &~\text{if the degree of}~f(x) > g(x) \end{cases} \end{aligned}$$
Difference of Square Root (Linear)
$$\displaystyle \lim_{x \to \infty} (\sqrt{ax+b}- \sqrt{cx + d}) = \begin{cases} \infty, &~\text{if}~a > c \\ 0, &~\text{if}~a = c \\-\infty, &~\text{if}~a < c \end{cases}$$Difference of Square Root (Quadratic)
$\begin{aligned} & \displaystyle \lim_{x \to \infty} (\sqrt{ax^2+bx+c}-\sqrt{px^2+qx+r}) \\ & = \begin{cases} \infty,~\text{if}~a > p \\ \dfrac{b-q} {2\sqrt{a}},~\text{if}~ a = p \\-\infty,~\text{if}~a < p \end{cases} \end{aligned}$
Difference of Cube Root
$$\displaystyle \lim_{x \to \infty} (\sqrt[3]{ax^3+bx^2+cx+d}-\sqrt[3]{ax^3+px^2+qx+r}) = \dfrac{b-p}{3\sqrt[3]{a^2}}$$

Today Quote

Mathematics is the language with which god has written the universe.

Multiple Choice Section

Problem Number 1
The value of $\displaystyle \lim_{x \to \infty} \dfrac{2x^3+3x^2-5x+4}{2x^4- 4x^2 + 9}$ is $\cdots \cdot$
A. $-\infty$                 C. $\infty$              E. $2$
B. $0$                      D. $1$

Solution

Formal Approach:
$\begin{aligned} & \displaystyle \lim_{x \to \infty} \dfrac{2x^3+3x^2-5x+4}{2x^4-4x^2 + 9} \\ & = \lim_{x \to \infty} \dfrac{\dfrac{2x^3}{x^4}+\dfrac{3x^2}{x^4}-\dfrac{5x} {x^4}+\dfrac{4}{x^4}}{\dfrac{2x^4}{x^4}- \dfrac{4x^2}{x^4} + \dfrac{9} {x^4}} \\ & = \dfrac{0-0-0+0}{2-0+0} = 0 \end{aligned}$
Quick Approach:
Notice that the numerator and denominator are polynomial functions. You may see that the power of leading term in numerator is $3$, while in the denominator is $4$. By applying the theorems noted above, we conclude that the limit tends to $0$ since $3 < 4$.
(Answer B)

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Problem Number 2
The value of $\displaystyle \lim_{x \to \infty} \dfrac{2x^3+3x^2+7}{x^2+3x+4}$ is $\cdots \cdot$
A. $\infty$               C. $-\infty$              E. $\dfrac{1}{2}$
B. $0$                  D. $2$          

Solution

Formal Approach:
Divide each term in numerator and denominator by $x^3$. 
$$\begin{aligned} \displaystyle \lim_{x \to \infty} \dfrac{2x^3+3x^2+7}{x^2+3x+4} & = \lim_{x \to \infty} \dfrac{\dfrac{2x^3}{x^3} + \dfrac{3x^2}{x^3}+\dfrac{7}{x^3}} {\dfrac{x^2}{x^3}+\dfrac{3x} {x^3}+\dfrac{4}{x^3}} = \infty \end{aligned}$$Quick Approach:
Notice that the numerator and denominator are polynomial functions. You may see that the power of leading term in numerator is $3$, while in the denominator is $2$. By applying the theorems noted above, we conclude that the limit tends to $\infty$ since $3 > 2$. 
Thus, the value of $\boxed{\displaystyle \lim_{x \to \infty} \dfrac{2x^3+3x^2+7}{x^2+3x+4} = \infty}$
(Answer A)

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Problem Number 3
The value of $\displaystyle \lim_{x \to \infty} \left(3-x+\dfrac{x^2-2x} {x+5}\right)$ is $\cdots \cdot$
A. $-4$                  C. $-2$               E. $\infty$
B. $-3$                  D. $0$        

Solution

$\begin{aligned} & \displaystyle \lim_{x \to \infty} \left(3-x+\dfrac{x^2-2x} {x+5}\right) \\ & = \lim_{x \to \infty} \left(\dfrac{(3-x)(x+5)} {x+5}+\dfrac{x^2-2x} {x+5}\right) \\ & = \lim_{x \to \infty} \dfrac{(3x + 15-\cancel{x^2}-5x)+(\cancel{x^2}-2x)}{x+5} \\ & = \lim_{x \to \infty} \dfrac{-4x + 15}{x + 5} = \dfrac{-4}{1} =-4 \end{aligned}$
Thus, the value of  $\boxed{\displaystyle \lim_{x \to \infty} \left(3-x+\dfrac{x^2-2x} {x+5}\right) =-4}$
(Answer A) 

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Problem Number 4
If $f(x) = x + \dfrac{x^2}{\sqrt{x^2-2x}}$, then $\displaystyle \lim_{x \to \infty} \dfrac{f(x)} {x} = \cdots \cdot$
A. $-2$                C. $1$                 E. $\infty$
B. $0$                    D. $2$       

Solution

Given that 
$\dfrac{f(x)} {x} = \dfrac{x + \dfrac{x^2}{\sqrt{x^2-2x}}} {x} = 1 + \dfrac{x} {\sqrt{x^2-2x}}$
Hence, we will have
$\begin{aligned} \displaystyle \lim_{x \to \infty} \dfrac{f(x)} {x} & = \lim_{x \to \infty} \left(1 + \dfrac{x} {\sqrt{x^2-2x}}\right) \\ & = \lim_{x \to \infty} \left(1 + \dfrac{\dfrac{x} {x}} {\sqrt{\dfrac{x^2-2x} {x^2}}}\right) \\ & = 1 + \dfrac{1}{\sqrt{1+0}} = 2 \end{aligned}$
Thus, the value of $\displaystyle \lim_{x \to \infty} \dfrac{f(x)} {x}$ is $\boxed{2}$
(Answer D)

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Problem Number 5
The value of $\displaystyle \lim_{x \to \infty} \dfrac{\sqrt{18x^2-x+1}-3x} {\sqrt{x^2+2x}}$ is $\cdots \cdot$
A. $\sqrt{15}$                               D. $3$
B. $3(\sqrt{2}-1)$                    E. $4,5$
C. $3(\sqrt{2}+1)$

Solution

Divide each term by the variable in leading term, that is $x$. 
$\begin{aligned}&  \displaystyle \lim_{x \to \infty} \dfrac{\sqrt{18x^2-x+1}-3x} {\sqrt{x^2+2x}}\\  & = \lim_{x \to \infty} \dfrac{\dfrac{1}{x}\sqrt{18x^2-x+1}- \dfrac{3x} {x}} {\dfrac{1}{x} \sqrt{x^2+2x}} \\ & = \lim_{x \to \infty} \dfrac{\dfrac{\sqrt{18x^2-x+1}} {x^2}- \dfrac{3x} {x}} {\dfrac{\sqrt{x^2+2x}} {x^2}} \\ & = \dfrac{\sqrt{18-0+0}- 3}{\sqrt{1+0}} \\ & = \dfrac{\sqrt{18}-3}{1} = 3\sqrt{2}-3 = 3(\sqrt{2}- 1) \end{aligned}$
Thus, the value of $\boxed{\displaystyle \lim_{x \to \infty} \dfrac{\sqrt{18x^2-x+1}-3x} {\sqrt{x^2+2x}} = 3(\sqrt{2}-1)}$
(Answer B)

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Problem Number 6
The value of $\displaystyle \lim_{x \to \infty} (\sqrt{x-\sqrt{x}}- \sqrt{x + \sqrt{x}})$ is $\cdots \cdot$
A. $0,5$
B. $1$
C. $-1$
D. $0$
E. $\text{the limi}\text{t doesn’t exist}$                       

Solution

Multiply both top and bottom by the conjugate. 
$$\begin{aligned} & \displaystyle \lim_{x \to \infty} (\sqrt{x-\sqrt{x}}-\sqrt{x + \sqrt{x}}) \\ & = \displaystyle \lim_{x \to \infty} (\sqrt{x-\sqrt{x}}-\sqrt{x + \sqrt{x}}) \times \dfrac{\sqrt{x-\sqrt{x}} + \sqrt{x + \sqrt{x}}}{\sqrt{x- \sqrt{x}} + \sqrt{x + \sqrt{x}}} \\ & = \lim_{x \to \infty} \dfrac{(x-\sqrt{x})-(x+\sqrt{x})} {\sqrt{x-\sqrt{x}} + \sqrt{x + \sqrt{x}}} \\ & =\lim_{x \to \infty} \dfrac{-2\sqrt{x}} {\sqrt{x-\sqrt{x}} + \sqrt{x + \sqrt{x}}} \end{aligned}$$Divide each term by $\sqrt{x}$. 
$\displaystyle \begin{aligned} & \lim_{x \to \infty} \dfrac{\dfrac{-2\sqrt{x}} {\sqrt{x}}}{\dfrac{\sqrt{x-\sqrt{x}}} {\sqrt{x}} + \dfrac{\sqrt{x + \sqrt{x}}} {\sqrt{x}}} \\ & = \lim_{x \to \infty} \dfrac{-2}{\sqrt{1-\dfrac{1}{\sqrt{x}}} + \sqrt{1+\dfrac{1}{\sqrt{x}}}} \\ & = \dfrac{-2}{\sqrt{1-0} + \sqrt{1+0}} \\ & = \dfrac{-2}{1+1} =-1 \end{aligned}$
Thus, the value of $\boxed{\displaystyle \lim_{x \to \infty} (\sqrt{x-\sqrt{x}}-\sqrt{x + \sqrt{x}}) =-1}$ 
(Answer C)

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Problem Number 7
The value of $\displaystyle \lim_{x \to \infty} x(\sqrt{x^2+1}-x)$ is $\cdots \cdot$
A. $-\dfrac14$                  C. $0$                 E. $\dfrac12$
B. $-\dfrac12$                  D. $\dfrac14$

Solution

Multiply by the conjugate of the expression involving square root form.
$\begin{aligned} & \displaystyle \lim_{x \to \infty} x(\sqrt{x^2+1}-x) \\ & = \lim_{x \to \infty} x(\sqrt{x^2+1}-x) \times \dfrac{\sqrt{x^2+1} + x} {\sqrt{x^2+1} + x} \\ & = \lim_{x \to \infty} \dfrac{x(x^2+1-x^2)} {\sqrt{x^2+1}+x} \\ & = \lim_{x \to \infty} \dfrac{x+1}{\sqrt{x^2+1}+x} \end{aligned}$
Divide each term with the variable of leading term, that is $x$. 
$\begin{aligned} \lim_{x \to \infty} \dfrac{x+1}{\sqrt{x^2+1}+x}  & = \lim_{x \to \infty} \dfrac{\dfrac{x+1}{x}}{\sqrt{\dfrac{x^2+1}{x^2}}+\dfrac{x} {x} } \\ & = \dfrac{1 + 0}{\sqrt{1+0} + 1} = \dfrac{1}{2} \end{aligned}$
Thus, the value of $\boxed{\displaystyle \lim_{x \to \infty} x(\sqrt{x^2+1}-x) = \dfrac{1}{2}}$
(Answer E)

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Problem Number 8
The value of $\displaystyle \lim_{x \to \infty} (\sqrt{x^4+2x^3+4x^2}-$ $\sqrt{x^4+2x^3-x^2})$ is $\cdots \cdot$
A. $0$                     C. $1$                   E. $\dfrac{5}{2}$
B. $\dfrac{1}{2}$                  D. $\dfrac{3}{2}$      

Solution

Multiply by the conjugate of the expression involving square root form.
$$\begin{aligned} & \displaystyle \lim_{x \to \infty} (\sqrt{x^4+2x^3+4x^2}-\sqrt{x^4+2x^3-x^2}) \\ & = \lim_{x \to \infty} (\sqrt{x^4+2x^3+4x^2}-\sqrt{x^4+2x^3-x^2}) \\ &  \times \dfrac{\sqrt{x^4+2x^3+4x^2}+\sqrt{x^4+2x^3-x^2}} {\sqrt{x^4+2x^3+4x^2}+\sqrt{x^4+2x^3-x^2}} \\ & = \lim_{x \to \infty} \dfrac{(x^4+2x^3+4x^2)-(x^4+2x^3-x^2)} {\sqrt{x^4+2x^3+4x^2}+\sqrt{x^4+2x^3-x^2}} \\ & = \lim_{x \to \infty} \dfrac{5x^2}{\sqrt{x^4+2x^3+4x^2}+\sqrt{x^4+2x^3-x^2}} \\ & \text{Divide each term by}~x^2 \\ & = \lim_{x \to \infty} \dfrac{\dfrac{5x^2}{x^2}} {\sqrt{\dfrac{x^4}{x^4}+\dfrac{2x^3}{x^4}+\dfrac{4x^2}{x^4}}+\sqrt{\dfrac{x^4}{x^4}+\dfrac{2x^3}{x^4}-\dfrac{x^2}{x^4}}} \\ & = \dfrac{5}{\sqrt{1+0+0} + \sqrt{1+0-0}} = \dfrac{5}{2} \end{aligned}$$Thus, the value of the limit is $\boxed{\dfrac{5}{2}}$
(Answer E)

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Problem Number 9
The value of $\displaystyle \lim_{x \to \infty} (\sqrt{9x^2+5x+5}-$ $\sqrt{9x^2-7x-4})$ is $\cdots \cdot$
A. $0$                     C. $1$                E. $3$
B. $\dfrac{1}{3}$                   D. $2$          

Solution

Apply the following formula.
$$\boxed{\displaystyle \lim_{x \to \infty} (\sqrt{ax^2+bx+c}-\sqrt{ax^2+px+q}) = \dfrac{b-p} {2\sqrt{a}} }$$In this case, it is given that
$a = 9, b = 5, p =-7$
Hence, we will have
$\begin{aligned} & \displaystyle \lim_{x \to \infty} (\sqrt{9x^2+5x+5}-\sqrt{9x^2-7x-4}) \\ & = \dfrac{5-(-7)} {2\sqrt{9}} = \dfrac{12}{6} = 2 \end{aligned}$
Thus, the value of the limit is $\boxed{2}$
(Answer D)

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Problem Number 10
The value of $\displaystyle \lim_{x \to \infty} (3x+1-\sqrt{9x^2+4x-7})$ is $\cdots \cdot$
A. $9$                  C. $3$                E. $\dfrac{1}{9}$
B. $6$                  D. $\dfrac{1}{3}$         

Solution

Notice that 
$3x+1 = \sqrt{(3x+1)^2} = \sqrt{9x^2+6x+1}$
can be applied since $x$ tends to infinity (the value must be positive at this rate).
Therefore, by applying this formula:
$$\boxed{\displaystyle \lim_{x \to \infty} (\sqrt{ax^2+bx+c}-\sqrt{ax^2+px+q}) = \dfrac{b-p} {2\sqrt{a}} }$$(Diketahui: $a = 9, b = 6, p = 4$)
we have
$$\displaystyle \lim_{x \to \infty} (3x+1)-\sqrt{9x^2+4x-7}) = \dfrac{6-4}{2\sqrt{9}} = \dfrac{1}{3}$$Thus, the value of the limit is $\boxed{\dfrac{1}{3}}$
(Answer D)

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Problem Number 11
The value of $\displaystyle \lim_{x \to \infty} (\sqrt{x^2+3x+2}-x+2)$ is $\cdots \cdot$
A. $5$                     C. $2,5$                E. $1$
B. $3,5$                 D. $1,5$        

Solution

You should notice that $-x+2$ can be rewritten:
$-(x-2) =-\sqrt{(x-2)^2} =-\sqrt{x^2-4x+4}$
Hence, the function can be rewritten into
$\displaystyle \lim_{x \to \infty} (\sqrt{x^2+3x+2}-(\sqrt{x^2-4x+4})$
Apply the following formula:
$$\boxed{\displaystyle \lim_{x \to \infty} (\sqrt{ax^2+bx+c}-\sqrt{ax^2+px+q}) = \dfrac{b-p} {2\sqrt{a}} }$$for $a = 1, b = 3, p =-4$, we have 
$\begin{aligned} & \displaystyle \lim_{x \to \infty} (\sqrt{x^2+3x+2}-(\sqrt{x^2-4x+4}) \\ & = \dfrac{3-(-4)} {2\sqrt{1}} = \dfrac{7}{2} = 3,5 \end{aligned}$
Thus, the value of the limit is $\boxed{3,5}$
(Answer B)

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Problem Number 12
The value of $\displaystyle \lim_{x \to \infty} ( \sqrt{(2x-1)(x+2)}-$ $(x\sqrt{2}+1))$ is $\cdots \cdot$
A. $3\sqrt{2}-4$
B. $\dfrac{3}{4}\sqrt{2}-1$
C. $\dfrac{3}{4}-\sqrt{2}$
D. $3-2\sqrt{2}$
E. $\sqrt{2}-1$

Solution

We will apply the following formula.
$\begin{aligned} & \displaystyle \lim_{x \to \infty} (\sqrt{ax^2+bx+c}-\sqrt{px^2+qx+r}) \\ & = \begin{cases} \infty,~\text{jika}~a > p \\ \dfrac{b-q} {2\sqrt{a}},~\text{jika}~ a = p \\-\infty,~\text{jika}~a < p \end{cases} \end{aligned}$
Hence, we will have
$\begin{aligned} & \displaystyle \lim_{x \to \infty} ( \sqrt{(2x-1)(x+2)}-(x\sqrt{2}+1)) \\ & = \lim_{x \to \infty} ((\sqrt{2x^2+3x-2}- \sqrt{2x^2})-1) \\ & = \dfrac{3-0}{2\sqrt{2}}-1 \\ & = \dfrac{3}{2\sqrt{2}}-1 \\ & = \dfrac{3}{4}\sqrt{2}-1 \end{aligned}$

Thus, the value of $$\boxed{\displaystyle \lim_{x \to \infty} ( \sqrt{(2x-1)(x+2)}- (x\sqrt{2}+1)) = \dfrac{3}{4}\sqrt{2}-1}$$(Answer B)

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Problem Number 13
The value of $\displaystyle \lim_{x \to \infty} [\sqrt{(x+a) (x+b)}-x]$ is $\cdots \cdot$
A. $\dfrac{a+b}{2}$                  D. $\dfrac{2}{a+b}$
B. $\dfrac{a-b}{2}$                   E. $\dfrac{a+b}{a-b}$
C. $\dfrac{ab}{2}$

Solution

$\begin{aligned}  & \displaystyle \lim_{x \to \infty} [\sqrt{(x+a) (x+b)}-x] \\ & = \lim_{x \to \infty} [\sqrt{x^2 + (a+b)x + ab}-\sqrt{x^2}] \\ & = \dfrac{(a+b)-0}{2\sqrt{1}} = \dfrac{a+b} {2} \end{aligned}$
Thus, the value of $\boxed{\displaystyle \lim_{x \to \infty} [\sqrt{(x+a) (x+b)}- x] = \dfrac{a+b} {2}}$
(Answer A)

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Problem Number 14
The value of $\displaystyle \lim_{x \to \infty} \sqrt{5(x-1)+2\sqrt{4x^2-23x-6}}$ is $\cdots \cdot$
A. $1$                   C. $-1$              E. $\infty$
B. $0$                   D. $-2$         

Solution

Apply one of radical’s property:
$\boxed{\sqrt{(a+b) \pm 2\sqrt{ab}} = \sqrt{a} \pm \sqrt b}$
and several fundamental properties of limit, we have
$$\begin{aligned} & \displaystyle \lim_{x \to \infty} \sqrt{5(x-1)+2\sqrt{4x^2-23x-6}} \\ & = \lim_{x \to \infty} \sqrt{[(4x + 1)+(x- 6)] + 2\sqrt{(4x+1)(x-6)}} \\ & = \lim_{x \to \infty} (\sqrt{4x+1} + \sqrt{x-6}) \\ & = \infty \end{aligned}$$The explanation for the last step: Since $x$ tends to infinity, the value of both $\sqrt{4x+1}$ and $\sqrt{x-6}$ will be getting bigger and bigger, so if we add them, the result must be infinity.
(Answer E)

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Problem Number 15
The value of $\displaystyle \lim_{x \to \infty} 2x\left(\sqrt{9+\dfrac{10}{x}}-3\right) $ $= \cdots \cdot$
A. $\dfrac{10}{3}$                      D. $\dfrac{5}{3}\sqrt{2}$
B. $-\dfrac{10}{3}$                   E. $-\dfrac{5}{3}\sqrt{2}$
C. $\dfrac{5}{3}$ 

Solution

By applying rationalizing method, we have
$$\begin{aligned} & \displaystyle \lim_{x \to \infty} 2x\left(\sqrt{9+\dfrac{10}{x}}- 3\right) \\ & = \lim_{x \to \infty} 2x\left(\sqrt{9+\dfrac{10}{x}}-3\right) \times \dfrac{\sqrt{9+\dfrac{10}{x}} + 3}{\sqrt{9+\dfrac{10}{x}} + 3} \\ & = \lim_{x \to \infty} \dfrac{2x \cdot \left(9+ \dfrac{10}{x}-3^2\right)}{\sqrt{9+\dfrac{10}{x}} + 3} \\ & = \lim_{x \to \infty} \dfrac{2 \cdot 10}{\sqrt{9+\dfrac{10}{x}} + 3} \\ & = \dfrac{20}{\sqrt{9+0} + 3} = \dfrac{10}{3} \end{aligned}$$Thus, the value of $\boxed{\displaystyle \lim_{x \to \infty} 2x\left(\sqrt{9+\dfrac{10}{x}}-3\right) = \dfrac{10}{3}}$ 
(Answer A)

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Problem Number 16
The value of $\displaystyle \lim_{x \to \infty} \dfrac{(x-2)\sqrt{(x+2) + \sqrt{4x}}}{x\sqrt{2x}-2\sqrt{x} + 2\sqrt{2}}$ $= \cdots \cdot$
A. $-\infty$                  C. $0$                 E. $1$
B. $-\dfrac12\sqrt2$              D. $\dfrac12\sqrt2$

Solution

Notice that
$\displaystyle \lim_{x \to \infty} \dfrac{(x-2)\sqrt{(x+2) + \sqrt{4x}}}{x\sqrt{2x}- 2\sqrt{x} + 2\sqrt{2}}$
can be expressed as
$\displaystyle \lim_{x \to \infty} \sqrt{\left(\dfrac{(x-2)\sqrt{(x+2) + \sqrt{4x}}}{x\sqrt{2x}-2\sqrt{x} + 2\sqrt{2}}\right)^2}$
Hence, we have
$\displaystyle \lim_{x \to \infty} \sqrt{\dfrac{(x^2-4x + 4)(x + 2 + \sqrt{4x})}{(\sqrt{2}x^{\frac32}-2x^{\frac12} + 2\sqrt{2})^2}}$
Observe the leading term in both top and bottom only.
$\displaystyle \lim_{x \to \infty} \sqrt{\dfrac{x^3 + \cdots}{2x^3 + \cdots}}$
Divide each term by $x^3$.
$\displaystyle \lim_{x \to \infty} \sqrt{\dfrac{\dfrac{x^3}{x^3} + \cdots}{\dfrac{2x^3}{x^3} + \cdots}}$
Lastly, apply limit at infinity’s properties to get
$\sqrt{\dfrac{1 + 0}{2 + 0}} = \dfrac{1}{\sqrt{2}} = \dfrac{1}{2}\sqrt{2}$
Thus, the value of $\boxed{\displaystyle \lim_{x \to \infty} \dfrac{(x-2)\sqrt{(x+2) + \sqrt{4x}}}{x\sqrt{2x}-2\sqrt{x} + 2\sqrt{2}} = \dfrac{1}{2}\sqrt{2}}$
(Answer D)

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Problem Number 17
The value of $\displaystyle \lim_{x \to \infty} \dfrac{3^{x+1} + 2^x-3}{3^{x+2}-2^{x-1} + 4} = \cdots \cdot$
A. $1$                       C. $\dfrac13$                  E. $\dfrac16$
B. $\dfrac12$                      D. $\dfrac14$        

Solution

$\begin{aligned} & \displaystyle \lim_{x \to \infty} \dfrac{3^{x+1} + 2^x-3}{3^{x+2}-2^{x-1} + 4} \\ & = \lim_{x \to \infty} \dfrac{3^{x+1} + 2^x-3}{3^{x+2}-2^{x-1} + 4} \times \dfrac{\frac{1}{3^{x+2}}} {\frac{1}{3^{x+2}}} \\ & = \lim_{x \to \infty} \dfrac{\frac13 + \frac{2^x} {3^{x+2}}-\dfrac{3}{3^{x+2}}} {1-\frac{2^{x-1}}{3^{x+2}} + \dfrac{4}{3^{x+2}}} \\ & = \dfrac{\frac13 + 0-0}{1-0 + 0} = \dfrac13 \end{aligned}$
Jadi, The value of $\boxed{\displaystyle \lim_{x \to \infty} \dfrac{3^{x+1} + 2^x-3}{3^{x+2}-2^{x-1} + 4} = \dfrac13}$
(Answer C)

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Problem Number 18 
The value of $\displaystyle \lim_{x \to \infty} \left(\sqrt[3]{8^x+3^x}- \sqrt{4^x-2^x}\right) = \cdots \cdot$
A. $-2$                   C. $-\dfrac12$                E. $\dfrac14$
B. $-1$       
            D. $\dfrac12$        

Solution

Factorize $2^x$ out of the radical form,
$$\begin{aligned} & \displaystyle \lim_{x \to \infty} \left(\sqrt[3]{8^x+3^x}- \sqrt{4^x-2^x}\right) \\ & = \lim_{x \to \infty} \left(\sqrt[3]{8^x\left(1 + \left(\dfrac38\right)^x\right)}- \sqrt{4^x\left(1- \left(\dfrac12\right)^x\right)}\right) \\ & = \lim_{x \to \infty} \left(2^x \sqrt[3]{1 + \left(\dfrac38\right)^x}- 2^x \sqrt{1- \left(\dfrac12\right)^x}\right) \end{aligned}$$Next, you should apply Binomial Approximation (Binomial Approach) to have 

$$\begin{aligned} & \lim_{x \to \infty} \left(2^x \left(1 + \dfrac13\left(\dfrac38\right)^x\right)-2^x \left(1-\dfrac12\left(\dfrac12\right)^x\right)\right) \\ & = \lim_{x \to \infty} \left(\cancel{2^x} + \dfrac13\left(\dfrac68\right)^x-\cancel{2^x} + \dfrac12\right) \\ & = \dfrac13(0) + \dfrac12 = \dfrac12 \end{aligned}$$Thus, the value of $\boxed{\displaystyle \lim_{x \to \infty} \left(\sqrt[3]{8^x+3^x}- \sqrt{4^x-2^x}\right) = \dfrac12}$
(Answer D)

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Problem Number 19
The value of $$\displaystyle \lim_{n \to \infty} \left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)\cdots\left(1-\dfrac{1}{n^2}\right) = \cdots \cdot$$
A. $\infty$                  C. $1$                 E. $\dfrac14$
B. $2$                  D. $\dfrac12$        

Solution

Simplify the function first by applying a factorization $a^2-b^2=(a+b)(a-b)$ and telescoping technique.
$$\begin{aligned} & \left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)\cdots\left(1-\dfrac{1}{n^2}\right) \\ & = \left(1-\dfrac12\right)\left(1+\dfrac12\right) \left(1-\dfrac13\right)\left(1+\dfrac13\right) \\ & \left(1-\dfrac14\right)\left(1+\dfrac14\right) \cdots \left(1-\dfrac{1}{n}\right)\left(1+\dfrac{1}{n}\right) \\ & = \dfrac12 \cdot \cancel{\dfrac32 \cdot \dfrac23 \cdot \dfrac43 \cdot \dfrac34 \cdot \dfrac54 \cdots \dfrac{n-1}{n}} \cdot \dfrac{n+1}{n} \\ & = \dfrac12 \cdot \dfrac{n+1}{n} = \dfrac{n+1}{2n} \end{aligned}$$Hence, we have
$\begin{aligned} \displaystyle \lim_{n \to \infty} \dfrac{n+1}{2n} & = \lim_{n \to \infty} \dfrac{\dfrac{n}{n}+\dfrac{1}{n}}{\dfrac{2n}{n}} \\ & = \dfrac{1+0}{2} = \dfrac12 \end{aligned}$
Thus, the value of
$\boxed{\begin{aligned} & \displaystyle \lim_{n \to \infty} \left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\\ & \left(1-\dfrac{1}{4^2}\right)\cdots\left(1-\dfrac{1}{n^2}\right) = \dfrac12 \end{aligned}}$

(Answer D)

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Problem Number 20
The value of $$\displaystyle \lim_{x \to \infty} \dfrac{\sqrt{9x^2+12x-1}-\sqrt{9x^2-24x+10}}{\sqrt[3]{x^3+8x^2+x}-\sqrt[3]{x^3-x^2+2x}} = \cdots \cdot$$
A. $\dfrac14$                  C. $1$                E. $3$
B. $\dfrac12$                  D. $2$         

Solution

Recall the following formulas.
$$\boxed{\begin{aligned} & \displaystyle \lim_{x \to \infty} (\sqrt{ax^2+bx+c}-\sqrt{ax^2+px+q})= \dfrac{b-p}{2\sqrt{a}} \\ & \lim_{x \to \infty} (\sqrt[3]{ax^3+bx^2+cx+d}-\sqrt[3]{ax^3+px^2+qx+r}) = \dfrac{b-p}{3\sqrt[3]{a^2}} \end{aligned}}$$Hence, we have
$\begin{aligned} & \displaystyle \lim_{x \to \infty} \dfrac{\sqrt{9x^2+12x-1}-\sqrt{9x^2-24x+10}}{\sqrt[3]{x^3+8x^2+x}-\sqrt[3]{x^3-x^2+2x}}\\ & = \dfrac{\dfrac{12-(-24)}{2\sqrt9}}{\dfrac{8-(-1)}{3\sqrt[3]{1^2}}} = \dfrac{\dfrac{36}{6}}{\dfrac{9}{3}} = \dfrac{6}{3} = 2 \end{aligned}$
Thus, the value of $$\boxed{\displaystyle \lim_{x \to \infty} \dfrac{\sqrt{9x^2+12x-1}-\sqrt{9x^2-24x+10}}{\sqrt[3]{x^3+8x^2+x}-\sqrt[3]{x^3-x^2+2x}} = 2}$$(Answer D)

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Problem Number 21
The value of $\displaystyle \lim_{x \to \infty} \left(3x + \sin \dfrac{1}{x} \right)$ is $\cdots \cdot$
A. $0$                   C. $4$                E. $\infty$
B. $3$                   D. $5$

Solution

Suppose $y = \dfrac{1}{x}$, which is equivalent to $x = \dfrac{1}{y}$. If $x \to \infty$, then $y \to 0$, so we have
$\begin{aligned} \displaystyle \lim_{x \to \infty} \left(3x + \sin \dfrac{1}{x} \right) & = \lim_{y \to 0} \left(\dfrac{3}{y} + \sin y \right) \\ & = \lim_{y \to 0} \dfrac{3}{y} + \lim_{y \to 0} \sin y \\ & = \infty + 0 = \infty \end{aligned}$
Thus, the value of $\boxed{\displaystyle \lim_{x \to \infty} \left(3x + \sin \dfrac{1}{x} \right) = \infty}$
(Answer E)

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Problem Number 22
The value of $\displaystyle \lim_{x \to \infty} \left(2 + \cos \dfrac{4}{x}\right)$ is $\cdots \cdot$
A. $0$                 C. $2$                E. $\infty$
B. $1$                 D. $3$

Solution

Suppose $y = \dfrac{1}{x}$. If $x \to \infty$, then $y \to 0$, so we have
$\begin{aligned} \displaystyle \lim_{x \to \infty} \left(2 + \cos \dfrac{4}{x}\right) & = \lim_{y \to 0} (2 + \cos 4y) \\ & = 2 + \cos 0 = 3 \end{aligned}$
Thus, the value of $\boxed{\displaystyle \lim_{x \to \infty} \left(2 + \cos \dfrac{4}{x}\right) = 3}$
(Answer D)

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Problem Number 23
The value of $\displaystyle \lim_{x \to \infty} \left(\tan \dfrac{1}{x}-x\right)$ is $\cdots \cdot$
A. $-\infty$                  C. $1$                 E. $\infty$
B. $0$                      D. $2$

Solution

Suppose $y = \dfrac{1}{x}$, which is equivalent to $x = \dfrac{1}{y}$. If $x \to \infty$, then $y \to 0$, so we have
$\begin{aligned} \displaystyle \lim_{x \to \infty} \left(\tan \dfrac{1}{x}-x\right) & = \lim_{y \to 0} \left(\tan y-\dfrac{1}{y} \right) \\ & = \tan 0- \infty =-\infty \end{aligned}$
Thus, the value of $\boxed{\displaystyle \lim_{x \to \infty} \left(\tan \dfrac{1}{x}- x\right) =-\infty}$
(Answer A)

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Problem Number 24
The value of $\displaystyle \lim_{x \to \infty} \sin \left(\dfrac{1}{x}-\dfrac{4 \pi} {3}\right)$ is $\cdots \cdot$
A. $-\infty$             C. $\dfrac12\sqrt2$              E. $1$
B. $0$                  D. $\dfrac12\sqrt3$

Solution

Suppose $y = \dfrac{1}{x}$. If $x \to \infty$, then $y \to 0$, so we have
$\begin{aligned} \displaystyle \lim_{x \to \infty} \sin \left(\dfrac{1}{x}-\dfrac{4 \pi} {3}\right) & = \lim_{y \to 0} \sin \left(y- \dfrac{4 \pi} {3}\right) \\ & = \sin \left(0-\dfrac{4 \pi} {3}\right) \\ & =-\sin 240^{\circ} = \dfrac{1}{2}\sqrt{3} \end{aligned}$
Thus, the value of $\boxed{\displaystyle \lim_{x \to \infty} \sin \left(\dfrac{1}{x}- \dfrac{4 \pi} {3}\right) = \dfrac{1}{2}\sqrt{3}}$
(Answer D)

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Problem Number 25
The value of $\displaystyle \lim_{x \to \infty} \left(\sin \left(\dfrac{1}{x}- \dfrac{6 \pi} {7}\right)-5x\right)$ is $\cdots \cdot$
A. $-\infty$                C. $0$               E. $\infty$
B. $-1$                 D. $1$

Solution

Suppose $y = \dfrac{1}{x}$, which is equivalent to $x = \dfrac{1}{y}$.
If $x \to \infty$, then $y \to 0$, so we have
$\begin{aligned} & \left(\sin \left(\dfrac{1}{x}-\dfrac{6 \pi} {7}\right)-5x\right) \\ & = \lim_{y \to 0} \left(\sin \left(y-\dfrac{6 \pi}{7}\right)-\dfrac{5}{y}\right) \\ & = \lim_{y \to 0} \sin \left(y-\dfrac{6 \pi}{7}\right)- \lim_{y \to 0} \dfrac{5}{y} \\ & =-\sin \dfrac{6 \pi}{7}-\infty =-\infty \end{aligned}$
Thus, the value of $\boxed{\displaystyle \lim_{x \to \infty} \left(\sin \left(\dfrac{1}{x}-\dfrac{6 \pi} {7}\right)-5x\right) =-\infty}$
(Answer A)

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Problem Number 26
The value of $\displaystyle \lim_{y \to \infty} \sqrt{6y} \cos \dfrac{3}{\sqrt{y}} \sin \dfrac{5}{\sqrt{y}}$ is $\cdots \cdot$
A. $0$                   C. $2\sqrt6$              E. $\infty$
B. $\sqrt6$                D. $5\sqrt6$

Solution

Suppose $x = \dfrac{1}{\sqrt{y}}$, which is equivalent to $\sqrt{y} = \dfrac{1}{x}$. 
If $y \to \infty$, then $x \to 0$, so we have
$\begin{aligned} & \displaystyle \lim_{y \to \infty} \sqrt{6y} \cos \dfrac{3}{\sqrt{y}} \sin \dfrac{5}{\sqrt{y}} \\ & = \lim_{x \to 0} \dfrac{\sqrt{6}} {x} \cos 3x \sin 5x \\ & = \lim_{x \to 0} \dfrac{\sin 5x} {x} \cdot \sqrt{6} \cdot \cos 3x \\ & = 5 \cdot \sqrt{6} \cos 0 \\ & = 5\sqrt{6} \end{aligned}$
Thus, the value of $\boxed{\displaystyle \lim_{y \to \infty} \sqrt{6y} \cos \dfrac{3}{\sqrt{y}} \sin \dfrac{5}{\sqrt{y}} = 5\sqrt{6}}$
(Answer D)

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Problem Number 27
The value of $\displaystyle \lim_{x \to \infty} \dfrac{1-\cos \dfrac{4}{x}} {\dfrac{1}{x} \cdot \tan \dfrac{3}{x}}$ is $\cdots \cdot$
A. $\dfrac13$                C. $\dfrac43$                 E. $\infty$
B. $\dfrac23$                D. $\dfrac83$

Solution

Suppose $y= \dfrac{1}{x}$. 
If $x \to \infty$, then $y \to 0$, so we have
$\begin{aligned}  \displaystyle \lim_{x \to \infty} \dfrac{1-\cos \dfrac{4}{x}} {\dfrac{1}{x} \cdot \tan \dfrac{3}{x}} & = \lim_{y \to 0} \dfrac{1-\cos 4y} {y \cdot \tan 3y} \\ & = \lim_{y \to 0} \dfrac{1-(1-2 \sin^2 2y)} {y \cdot \tan 3y} \\ & = \lim_{y \to 0} \dfrac{2 \sin 2y \sin 2y} {y \cdot \tan 3y} \\ & = 2 \lim_{y \to 0} \dfrac{\sin 2y} {y} \cdot \dfrac{\sin 2y} {\tan 3y} \\ & = 2 \cdot 2 \cdot \dfrac{2}{3} = \dfrac{8}{3} \end{aligned}$
Thus, the value of $\boxed{\displaystyle \lim_{x \to \infty} \dfrac{1-\cos \dfrac{4}{x}} {\dfrac{1}{x} \cdot \tan \dfrac{3}{x}} = \dfrac{8}{3}}$
(Answer D)

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Problem Number 28
The value of $\displaystyle \lim_{x \to \infty} 2x^2\left(1-\cos \dfrac{6}{x} \right)$ is $\cdots \cdot$
A. $0$                 C. $12$               E. $36$
B. $6$                 D. $18$

Solution

Suppose $y = \dfrac{1}{x}$, which is equivalent to $x = \dfrac{1}{y}$. 
If $x \to \infty$, then $y \to 0$. 
Therefore, we have
$\begin{aligned} & \displaystyle \lim_{x \to \infty} 2x^2\left(1-\cos \dfrac{6}{x} \right) \\ & = \lim_{y \to 0} \dfrac{2}{y^2}(1-\cos 6y) \\ & = \lim_{y \to 0} \dfrac{2}{y^2}(1- \cos 6y) \times \dfrac{1+\cos 6y} {1+\cos 6y} \\ & = 2 \lim_{y \to 0} \dfrac{1-\cos^2 6y} {y^2(1 + \cos 6y)} \\ & = \lim_{y \to 0} \dfrac{\sin 6y} {y} \cdot \dfrac{\sin 6y} {y} \cdot \dfrac{1}{1+\cos 6y} \\ & = 6 \cdot 6 \cdot \dfrac{1}{1+ \cos 0} = 36 \cdot \dfrac12 = 18 \end{aligned}$
Thus, the value of $\boxed{\displaystyle \lim_{x \to \infty} 2x^2\left(1-\cos \dfrac{6}{x} \right) = 18}$
(Answer D)

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Problem Number 29
The value of $\displaystyle \lim_{\theta \to \infty} \dfrac{\sin^2 \theta} {\theta^2-5}$ is $\cdots \cdot$
A. $-\infty$
B. $0$
C. $1$
D. $\infty$
E. $\text{the limi}\text{t doesn’t exist}$

Solution

$$\begin{aligned} & \displaystyle \lim_{\theta \to \infty} \dfrac{\sin^2 \theta} {\theta^2-5} \\ & = \lim_{\theta \to \infty} \dfrac{\sin \theta} {\theta + \sqrt{5}} \times \dfrac{\sin \theta} {\theta-\sqrt{5}} \\ & = \lim_{\theta \to \infty} \dfrac{\sin \theta} {\theta} \times \dfrac{1}{1 + \dfrac{\sqrt{5}} {\theta}} \times \dfrac{\sin \theta} {\theta} \times \dfrac{1}{1-\dfrac{\sqrt{5}} {\theta}} \\ & = (0 \times 1 \times 0 \times 1) = 0 \end{aligned}$$Thus, the value of $\boxed{\displaystyle \lim_{\theta \to \infty} \dfrac{\sin^2 \theta} {\theta^2-5} = 0}$
(Answer B)

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Problem Number 30 
The value of $\displaystyle \lim_{y \to \infty} y \cdot \sin \dfrac{3}{y} \cdot \cos \dfrac{5}{y}$ is $\cdots \cdot$
A. $0$                  C. $2$              E. $4$
B. $1$                  D. $3$      

Solution

Suppose $x= \dfrac{1}{y}$. 
If $y \to \infty$, then $x \to 0$, so we have
$$\begin{aligned} \displaystyle \lim_{y \to \infty} y \cdot \sin \dfrac{3}{y} \cdot \cos \dfrac{5}{y} & = \lim_{x \to 0} \dfrac{1}{x} \cdot \sin 3x \cdot \cos 5x \\ & = \lim_{x \to 0} \dfrac{\sin 3x} {x} \cdot \cos 5x \\ & = 3 \cdot \cos 0 = 3 \end{aligned}$$Thus, the value of $\displaystyle \lim_{y \to \infty} y \cdot \sin \dfrac{3}{y} \cdot \cos \dfrac{5}{y}$ is $\boxed{3}$
(Answer D)

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Problem Number 31 
The value of $\displaystyle \lim_{x \to \infty} \dfrac{\sin \dfrac{3}{x}} {\left(1-\cos \dfrac{2}{x} \right) \cdot x^2 \cdot \sin \dfrac{1}{x}}$ is $\cdots \cdot$
A. $0$                     C. $1$               E. $3$
B. $\dfrac{2}{3}$                   D. $\dfrac{3}{2}$       

Solution

Suppose $y= \dfrac{1}{x}$, which is equivalent to $x = \dfrac{1}{y}$. 
If $x \to \infty$, then $y \to 0$, so we have
$\begin{aligned} & \displaystyle \lim_{x \to \infty} \dfrac{\sin \dfrac{3}{x}} {\left(1-\cos \dfrac{2}{x} \right) \cdot x^2 \cdot \sin \dfrac{1}{x}} \\ & = \lim_{y \to 0} \dfrac{\sin 3y} {(1-\cos 2y) \cdot \left(\dfrac{1}{y}\right)^2 \cdot \sin y} \\ & = \lim_{y \to 0} \dfrac{\sin 3y \cdot y^2}{(1-(1-2 \sin^2 y)) \cdot \sin y} \\ & = \dfrac12 \lim_{y \to 0} \dfrac{\sin 3y} {\sin y} \cdot \dfrac{y}{\sin y} \cdot \dfrac{y}{\sin y} \\ & = \dfrac12 \times 3 \times 1 \times 1 = \dfrac32 \end{aligned}$
Thus, the value of $\boxed{\displaystyle \lim_{x \to \infty} \dfrac{\sin \dfrac{3}{x}} {\left(1-\cos \dfrac{2}{x} \right) \cdot x^2 \cdot \sin \dfrac{1}{x}} = \dfrac32}$
(Answer D)

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Problem Number 32 (Soal SBMPTN Tahun 2017 Saintek Kode 167)
The value of $\displaystyle \lim_{x \to \infty} x \left(1- \cos \dfrac{1}{\sqrt{x}}\right) = \cdots \cdot$
A. $1$                  C. $\dfrac{1}{3}$                 E. $\dfrac{1}{5}$
B. $\dfrac{1}{2}$                D. $\dfrac{1}{4}$     

Solution

Suppose $y= \dfrac{1}{\sqrt{x}}$, which is equivalent to $x = \dfrac{1}{y^2}$. 
If $x \to \infty$, then $y \to 0$, so we have
$\begin{aligned} & \displaystyle \lim_{x \to \infty} x \left(1- \cos \dfrac{1}{\sqrt{x}}\right) \\ & = \lim_{y \to 0} \dfrac{1}{y^2} (1-\cos y) \\ & = \lim_{y \to 0} \dfrac{1-\cos y} {y^2} \times \dfrac{1+\cos y} {1 + \cos y} \\ & = \lim_{y \to 0} \dfrac{1-\cos^2 y} {y^2(1 + \cos y)} \\ & = \lim_{y \to 0} \dfrac{\sin y} {y} \cdot \dfrac{\sin y} {y} \cdot \dfrac{1}{1 + \cos y} \\ & = 1 \cdot 1 \cdot \dfrac{1}{1 + \cos 0} = \dfrac{1}{2} \end{aligned}$
Thus, the value of $\boxed{\displaystyle \lim_{x \to \infty} x \left(1- \cos \dfrac{1}{\sqrt{x}}\right) = \dfrac{1}{2}}$
Note:
We are applying a Pythagorean Identity here:
$$\boxed{\sin^2 x + \cos^2 x = 1 \iff 1- \cos^2 x = \sin^2 x}$$

(Answer B) 
 

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Problem Number 33 
The value of $\displaystyle \lim_{x \to \infty} 2x \cdot \tan \dfrac{1}{x} \cdot \sec \dfrac{2}{x} = \cdots \cdot$
A. $0$                 C. $2$              E. $4$
B. $1$                 D. $3$      

Solution

Suppose $y= \dfrac{1}{x}$, which is equivalent to $x = \dfrac{1}{y}$. 
If $x \to \infty$, then $y \to 0$, so we have
$\begin{aligned} & \lim_{x \to \infty} 2x \cdot \tan \dfrac{1}{x} \cdot \sec \dfrac{2}{x} \\ & = \lim_{y \to 0} \dfrac{2}{y} \cdot \tan y \cdot \sec 2y \\ & = 2 \lim_{y \to 0} \dfrac{\tan y} {y} \cdot \sec 2y \\ & = 2 \cdot 1 \cdot \sec 0 = 2 \end{aligned}$
Thus, the value of $\boxed{\lim_{x \to \infty} 2x \cdot \tan \dfrac{1}{x} \cdot \sec \dfrac{2}{x} = 2}$
(Answer C)

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Problem Number 34 
The value of $\displaystyle \lim_{x \to \infty} \dfrac{2x^2 \tan \left(\dfrac{1}{x} \right)-x \sin \left(\dfrac{1}{x} \right) + \dfrac{1}{x}} {x \cos \left(\dfrac{2}{x} \right)} = \cdots \cdot$
A. $2$                  C. $0$                 E. $-2$
B. $1$                  D. $-1$          

Solution

Suppose $y = \dfrac{1}{x}$, which is equivalent to $x = \dfrac{1}{y}$. 
If $x \to \infty$, then $y \to 0$, so we have
$\begin{aligned} & \displaystyle \lim_{x \to \infty} \dfrac{2x^2 \tan \left(\dfrac{1}{x} \right)-x \sin \left(\dfrac{1}{x} \right) + \dfrac{1}{x}} {x \cos \left(\dfrac{2}{x} \right)} \\ & = \lim_{y \to 0} \dfrac{2\left(\dfrac{1}{y} \right)^2 \tan y- \dfrac{1}{y} \sin y + y} {\dfrac{1}{y} \cos 2y}\color{red} { \times \dfrac{y}{y}} \\ & = \lim_{y \to 0} \dfrac{2\left(\dfrac{1}{y} \right) \tan y-\sin y + y^2} {\cos 2y} \\ & = \lim_{y \to 0} \dfrac{2 \cdot \dfrac{\tan y}{y}- \sin y + y^2} {\cos 2y} \\ & = \dfrac{2 \cdot 1- \sin 0 + 0^2}{\cos 0} \\ & = \dfrac{2-0}{1} = 2 \end{aligned}$
Thus, the value of $$\boxed{\displaystyle \lim_{x \to \infty} \dfrac{2x^2 \tan \left(\dfrac{1}{x} \right)-x \sin \left(\dfrac{1}{x} \right) + \dfrac{1}{x}} {x \cos \left(\dfrac{2}{x} \right)} = 2}$$(Answer A)

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Problem Number 35
The value of $\displaystyle \lim_{x \to \infty} \dfrac{(x^2-3) \cos x}{x^3+1} = \cdots \cdot$
A. $-1$                    D. tidak ada
B. $0$                       E. $\infty$
C. $1$

Solution

Notice that 
$-\dfrac{x^2-3}{x^3+1}\le \dfrac{(x^2-3)\cos x}{x^3+1}\le \dfrac{x^2-3}{x^3+1}$
since $-1 \le \cos x \leq 1$.
You should also note
$\begin{aligned} \lim_{x\to \infty}\left(-\dfrac{x^2-3}{x^3+1}\right) & =\lim _{x\to \infty}\left(-\dfrac{\dfrac{1}{x}-\dfrac{3}{x^3}}{1+\dfrac{1}{x^3}}\right) \\ & =-\dfrac{0}{1}=0 \end{aligned}$
and
$\begin{aligned} \lim_{x\to \infty}\left(\dfrac{x^2-3}{x^3+1}\right) & =\lim _{x\to \infty}\left(\dfrac{\dfrac{1}{x}-\dfrac{3}{x^3}}{1+\dfrac{1}{x^3}}\right) \\ & =\dfrac{0}{1}=0 \end{aligned}$
By applying squeeze theorem, we eventually have
$\boxed{\displaystyle \lim_{x \to \infty} \dfrac{(x^2-3) \cos x}{x^3+1}=0}$
(Answer B)

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Essay Section

Problem Number 1
Find the value of:
a. $\displaystyle \lim_{x \to \infty} (4x + 2)$
b. $\displaystyle \lim_{x \to \infty} (-x + 4)$
c. $\displaystyle \lim_{x \to \infty}-(3x^2 + 9)$

Solution

Answer a) 
$\displaystyle \lim_{x \to \infty} (4x + 2) = 4(\infty) + 2 = \infty + 2 = \infty$
Answer b) 
$\displaystyle \lim_{x \to \infty} (-x + 4) =-\infty + 4 =-\infty$
Answer c) 
$\begin{aligned} \displaystyle \lim_{x \to \infty}-(3x^2 + 9) & =-(3(\infty)^2 + 9) \\ & =-(\infty + 9) =-\infty \end{aligned}$

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Problem Number 2
Find the value of the following limits.
a. $\displaystyle \lim_{x \to \infty} \dfrac{3x^3-2x-10}{ 4x-2x^2-5x^3}$
b. $\displaystyle \lim_{x \to \infty} \dfrac{x^5-2x^4+x^3-3x^2+2x-7}{7-2x+3x^2-x^3+2x^4}$

Solution

Answer a) 
Given that both numerator and denominator have the same highest power, that is $x^3$. The coefficients of leading terms in numerator and denominator are $3$ and $-5$, respectively. Thus,
$\displaystyle \lim_{x \to \infty} \dfrac{3x^3-2x-10}{ 4x-2x^2-5x^3} =-\dfrac{3}{5}$
Answer b) 
Since the numerator has the higher power than the denominator, as $5 > 4$, we have
$\displaystyle \lim_{x \to \infty} \dfrac{x^5-2x^4+x^3-3x^2+2x-7}{7-2x+3x^2-x^3+2x^4} = \infty$

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Problem Number 3
Evaluate the value of the following limits.
a. $\displaystyle \lim_{x \to \infty} \dfrac{(1-2x)^3}{(x-1)(2x^2+x+1)}$
b. $\displaystyle \lim_{x \to \infty} \dfrac{(3x-2)^3}{(4x+2)^3}$

Solution

Expand and observe the leading terms only.
Answer a) 
$$\begin{aligned} \displaystyle \lim_{x \to \infty} \dfrac{(1-2x)^3}{(x-1)(2x^2+x+1)} & = \lim_{x \to \infty} \dfrac{-8x^3 + \cdots} {2x^3 + \cdots} \\ & = \dfrac{-8}{2} =-4 \end{aligned}$$Thus, the value of
$\boxed{\displaystyle \lim_{x \to \infty} \dfrac{(1-2x)^3}{(x-1)(2x^2+x+1)} =-4}$
Answer b) 
$\displaystyle \lim_{x \to \infty} \dfrac{(3x-2)^3}{(4x+2)^3} = \lim_{x \to \infty} \dfrac{27x^3 + \cdots} {64x^3 + \cdots} = \dfrac{27}{64}$

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Problem Number 4
Evaluate the value of:
a. $\displaystyle \lim_{x \to \infty} (\sqrt{x+5}-\sqrt{x-3})$
b. $\displaystyle \lim_{x \to \infty} (\sqrt{2x-7}-\sqrt{x+3})$
c. $\displaystyle \lim_{x \to \infty} (\sqrt{x+5}-\sqrt{2x-3})$

Solution

You should recall that
$$\boxed{\displaystyle \lim_{x \to \infty} (\sqrt{ax+b}- \sqrt{cx + d}) = \begin{cases} \infty, &~\text{jika}~a > c \\ 0, &~\text{jika}~a = c \\-\infty, &~\text{jika}~a < c \end{cases}}$$Answer a) 
Given: $a = 1$ and $c = 1$, sehingga $a = c$. This means, 
$\displaystyle \lim_{x \to \infty} (\sqrt{x+5}-\sqrt{x-3}) = 0$
Answer b) 
Given: $a = 2$ and $c = 1$, sehingga $a > c$. This means, 
$\displaystyle \lim_{x \to \infty} (\sqrt{2x-7}-\sqrt{x+3}) = \infty$
Answer c) 
Given: $a = 1$ and $c = 2$, so $a < c$. This means,
$\displaystyle \lim_{x \to \infty} (\sqrt{x+5}-\sqrt{2x-3}) =-\infty$

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Problem Number 5
Evaluate the value of $\displaystyle \lim_{\theta \to-\infty} \dfrac{\pi \theta^5}{\theta^5- 5\theta^4}$
(Note: The Greek letter $\pi$ is read: pi, while $\theta$ is read: theta)

Solution

$$\begin{aligned} \displaystyle \lim_{\theta \to-\infty} \dfrac{\pi \theta^5}{\theta^5-5\theta^4} & = \lim_{\theta \to-\infty} \dfrac{\cancel{\theta^4}(\pi \theta)}{\cancel{\theta^4}(\theta-5)} \\ & = \lim_{\theta \to-\infty} \dfrac{\pi \theta} {\theta-5} \\ & = \lim_{\theta \to-\infty} \left(\dfrac{\pi(\theta-5)} {\theta-5} + \dfrac{5\pi} {\theta-5}\right) \\ & = \lim_{\theta \to-\infty} \left(\pi + \dfrac{5\pi} {\theta-5}\right) \\ & = \pi-0 = \pi \end{aligned}$$Jadi, The value of $\boxed{\displaystyle \lim_{\theta \to-\infty} \dfrac{\pi \theta^5}{\theta^5-5\theta^4} = \pi}$
Catatan: Tinjau bentuk $\dfrac{5\pi} {\theta-5}$. Apabila nilai $\theta$ semakin kecil menuju negatif tak hingga, maka penyebutnya juga akan semakin kecil, dan nilai pecahannya akan semakin mendekati $0$.

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Problem Number 6
Find the value of:

  1. $\displaystyle \lim_{x \to \infty} \left(\sqrt{(x+5)(4x+7)}-\sqrt{(x+3)(4x+7)}\right)$
  2. $\displaystyle \lim_{x \to \infty} (x- \sqrt{x^2-10x })$

Solution

Answer a) 
$$\begin{aligned} & \displaystyle \lim_{x \to \infty} (\sqrt{(x+5)(4x+7)}-\sqrt{(x+3)(4x+7)}) \\ & = \lim_{x \to \infty} (\sqrt{4x^2 + 27x + 35}- \sqrt{4x^2+19x+21}) \\ & = \dfrac{27-19}{2\sqrt{4}} = \dfrac{8}{4} = 2 \end{aligned}$$Thus, the value of $$\boxed{\displaystyle \lim_{x \to \infty} (\sqrt{(x+5)(4x+7)}- \sqrt{(x+3)(4x+7)}) = 2}$$Answer b) 
$$\begin{aligned} \displaystyle \lim_{x \to \infty} (x-\sqrt{x^2-10x }) & = \lim_{x \to \infty} (\sqrt{x^2}-\sqrt{x^2-10x}) \\ & = \dfrac{0-(-10)} {2\sqrt{1}} = \dfrac{10}{2} = 5 \end{aligned}$$Thus, the value of $\boxed{\displaystyle \lim_{x \to \infty} (x-\sqrt{x^2-10x }) = 5}$

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Problem Number 7
Find the value of $\displaystyle \lim_{x \to \infty} \dfrac{\sqrt{8x^2+1}} {x^2+4}$.

Solution

Divide each term by $x^4$ (highest power). 
$\begin{aligned} \displaystyle \lim_{x \to \infty} \dfrac{\sqrt{8x^2+1}} {x^2+4} & = \lim_{x \to \infty} \dfrac{\sqrt{\dfrac{8x^2}{x^4}+\dfrac{1}{x^4}}}{\dfrac{x^2}{x^2}+\dfrac{4}{x^2}} \\ & = \dfrac{\sqrt{0 + 0}} {1 + 0} = 0 \end{aligned}$
Thus, the value of $\boxed{\displaystyle \lim_{x \to \infty} \dfrac{\sqrt{8x^2+1}} {x^2+4} = 0}$

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Problem Number 8
Find the value of $\displaystyle \lim_{x \to \infty} (\sqrt{4x^2+8x}-\sqrt{x^2+1}-\sqrt{x^2+x})$

Solution

$$\begin{aligned} & \displaystyle \lim_{x \to \infty} (\sqrt{4x^2+8x}-\sqrt{x^2+1}- \sqrt{x^2+x}) \\ & = \lim_{x \to \infty} (2\sqrt{x^2+2x}- \sqrt{x^2+1}-\sqrt{x^2+x}) \\ & = \lim_{x \to \infty} ((\sqrt{x^2 + 2x}-\sqrt{x^2 + 1}) + (\sqrt{x^2+2x}-\sqrt{x^2 + x})) \\ & = \dfrac{2-0}{2\sqrt{1}} + \dfrac{2-1}{2\sqrt{1}} \\ & = \dfrac{2}{2} + \dfrac{1}{2} = \dfrac{3}{2} \end{aligned}$$Thus, the value of the limit is $\boxed{\dfrac32}$

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Problem Number 9
Evaluate the value of $\displaystyle \lim_{x \to \infty} \dfrac{\sqrt{3x^2-2x-1}} {x+2.000}$.

Solution

Divide each term in both numerator and denominator by $x$.
$\begin{aligned} & \displaystyle \lim_{x \to \infty} \dfrac{\sqrt{3x^2-2x-1}} {x+2.000} \\  & = \lim_{x \to \infty} \dfrac{\dfrac{1}{x} \sqrt{3x^2-2x-1}} {\dfrac{1}{x}\left(x+2.000\right)} \\ & = \lim_{x \to \infty} \dfrac{\sqrt{\dfrac{3x^2-2x-1}{x^2}}} {1 + \dfrac{2.000}{x}} \\ & = \lim_{x \to \infty} \dfrac{\sqrt{3-\dfrac{2}{x}- \dfrac{1}{x^2}}} {1 + \dfrac{2.000}{x}} \\ & = \dfrac{\sqrt{3-0-0}} {1 + 0} = \sqrt{3} \end{aligned}$
Thus, the value of $\displaystyle \lim_{x \to \infty} \dfrac{\sqrt{3x^2-2x-1}} {x+2.000}$ is $\boxed{\sqrt{3}}$

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Problem Number 10
Evaluate the value of $\displaystyle \lim_{x\to\infty}(\sqrt[7]{x^7+x^6}-\sqrt[7]{x^7-x^6})$.

Solution

Alternative I: Intuitive Approach
$$\begin{aligned} \displaystyle & \displaystyle \lim_{x\to\infty}(\sqrt[7]{x^7+x^6}-\sqrt[7]{x^7-x^6}) \\ = & \lim_{x\to\infty}\left(\sqrt[7] {\left (x+\dfrac17\right)^7+O(x^5)}-\sqrt[7]{\left(x-\dfrac17 \right)^7+O(x^5)}\right) \\ = & \lim_{x\to\infty} \left(\sqrt[7]{\left(x+\dfrac17\right)^7}-\sqrt[7]{\left(x-\dfrac17\right)^7}\right) \\ = &\lim_{x\to\infty}\left(\left(x+\dfrac17\right)- \left(x-\dfrac17\right)\right) \\ = & \lim_{x \to \infty} \left(\dfrac17 + \dfrac17\right) = \dfrac27 \end{aligned}$$Catatan: $O(x^5)$ denotes the fifth-degree polynomial as the exspansion of $\left(x+\dfrac{1}{7}\right)^7$. Because $x$ tends to infinity, the expression $\left(x+\dfrac{1}{7}\right)^7$ will getting bigger in value at a much faster rate, causing $O(x^5)$ can be ignored.
Alternative II: L’Hospital’s Rule
$$\begin{aligned} & \displaystyle \lim_{x\to\infty}(\sqrt[7]{x^7+x^6}-\sqrt[7]{x^7-x^6})\\ = & \lim_{x\to\infty} \left(\sqrt[7]{x^7\left(x + \dfrac1x\right)}-\sqrt[7]{x^7\left(x-\dfrac1x\right)}\right) \\ = & \lim_{x\to\infty} x\left(\sqrt[7]{1 + \dfrac1x}-\sqrt[7]{1-\dfrac1x}\right) \\ & \text{Misalkan}~x = \dfrac{1}{t} \\ = & \lim_{t \to 0} {{\sqrt[7]{1+t}-\sqrt[7]{1-t}}\over t} \\ \stackrel{\text{L’H}}{=} & \lim_{t \to 0} \left(\dfrac{1}{7}(1 + t)^{-\frac{6}{7}}(1)- \dfrac{1}{7}(1-t)^{-\frac{6}{7}}(-1)\right) \\ = &  \dfrac{1}{7}(1 + 0)^{-\frac{6}{7}} + \dfrac{1}{7}(1- 0)^{-\frac{6}{7}} \\ = & \dfrac{1}{7} + \dfrac{1}{7} =\dfrac27 \end{aligned}$$Thus, the value of $\displaystyle \lim_{x\to\infty}(\sqrt[7]{x^7+x^6}-\sqrt[7]{x^7-x^6})$ is $\boxed{\dfrac{2}{7}}$

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Problem Number 11
Evaluate the value of following limits.
a. $\displaystyle \lim_{x \to \infty} x \tan \dfrac{1}{x}$
b. $\displaystyle \lim_{x \to \infty} \dfrac{1}{x} \cot \dfrac{1}{x}$ 
c. $\displaystyle \lim_{x \to \infty} \dfrac{\csc \frac{1}{x}} {x} $

Solution

Answer a) 
Suppose $y = \dfrac{1}{x}$ which is equivalent to $x = \dfrac{1}{y}$. 
If $x \to \infty$, then $y \to 0$, so the limit can be rewritten as the following.
$\displaystyle \lim_{y \to 0} \dfrac{1}{y} \tan y = 1$
Hence, the value of $\boxed{\displaystyle \lim_{x \to 0} x \tan \dfrac{1}{x} = 1}$
Answer b) 
Recall that: $\boxed{\cot x = \dfrac{1}{\tan x}}$ 
Suppose $y = \dfrac{1}{x}$. 
If $x \to \infty$, then $y \to 0$, so the limit can be rewritten as the following.
$\displaystyle \lim_{y \to 0} y \cot y = \lim_{y \to 0} \dfrac{y} {\tan y} = 1$
Thus, the value of $\boxed{\displaystyle \lim_{x \to 0} \dfrac{1}{x} \cot \dfrac{1}{x} = 1}$
Answer c) 
Recall that: $\boxed{\csc x = \dfrac{1}{\sin x}}$
Suppose $y = \dfrac{1}{x}$ which is equivalent to $x = \dfrac{1}{x}$.
If $x \to \infty$, then $y \to 0$, so the limit can be rewritten as the following.
$\displaystyle \lim_{y \to 0} \dfrac{\csc y} {\frac{1}{y}} = \lim_{y \to 0} \dfrac{y} {\sin y} = 1$
Thus, the value of $\boxed{\displaystyle \lim_{x \to \infty} \dfrac{\csc \frac{1}{x}} {x} = 1}$  

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Problem Number 12
Evaluate the value of following limits.
a. $\displaystyle \lim_{x \to \infty} \tan \dfrac{5}{x} \csc \dfrac{2}{x}$
b. $\displaystyle \lim_{x \to \infty} \cot 3x^{-1} \sin x^{-1}$
c. $\displaystyle \lim_{x \to \infty} \dfrac{\cot \dfrac{1}{2x}} {\csc \dfrac{3}{x}}$

Solution

Answer a) 
Suppose $y = \dfrac{1}{x}$. 
If $x \to \infty$, then $y \to 0$. Hence, we will have
$\begin{aligned} \displaystyle \lim_{x \to \infty} \tan \dfrac{5}{x} \csc \dfrac{2}{x} & = \lim_{y \to 0} \tan 5y \csc 2y \\ & = \lim_{y \to 0} \dfrac{\tan 5y} {\sin 2y} = \dfrac{5}{2} \end{aligned}$
Thus, the value of $\displaystyle \lim_{x \to \infty} \tan \dfrac{5}{x} \csc \dfrac{2}{x} = \dfrac{5}{2}$
Answer b) 
Suppose $y = x^{-1}$. 
If $x \to \infty$, then $y \to 0$. Hence, we will have
$\begin{aligned} \displaystyle \lim_{x \to \infty} \cot 3x^{-1} \sin x^{-1} & = \lim_{y \to 0} \cot 3y \sin y \\ & = \lim_{y \to 0} \dfrac{\sin y} {\tan 3y} = \dfrac{1}{3} \end{aligned}$
Thus, the value of $\boxed{\displaystyle \lim_{x \to \infty} \cot 3x^{-1} \sin x^{-1} = \dfrac{1}{3}}$
Answer c) 
Suppose $y = \dfrac{1}{x}$.
If $x \to \infty$, then $y \to 0$. Hence, we will have
$\begin{aligned} \displaystyle \lim_{x \to \infty} \dfrac{\cot \dfrac{1}{2x}} {\csc \dfrac{3}{x}} & = \lim_{y \to 0} \dfrac{\cot \dfrac{1}{2}y} {\csc 3y} \\ & = \lim_{y \to 0} \dfrac{\sin 3y} {\tan \dfrac{1}{2}y} \\ & = \dfrac{3}{\dfrac{1}{2}} = 6 \end{aligned}$
Thus, the value of $\boxed{\displaystyle \lim_{x \to \infty} \dfrac{\cot \dfrac{1}{2x}} {\csc \dfrac{3}{x}} = 6}$

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Problem Number 13
Evaluate the following limits.
a. $\displaystyle \lim_{n \to \infty} \dfrac{5^n-10^n}{3^{2n}}$
b. $\displaystyle \lim_{n \to \infty} \dfrac{4+2^{2n}}{3^n-2}$
c. $\displaystyle \lim_{n \to \infty} \dfrac{(-3)^n}{2^n+1}$
d. $\displaystyle \lim_{n \to \infty} \dfrac{3^{n+1}+5^{n+1}+7^{n+1}}{3^n + 5^n + 7^n}$

Solution

Answer a)
Divide each term by $3^{2n}$, then apply the limit at infinity theorems.
$\begin{aligned} \displaystyle \lim_{n \to \infty} \dfrac{5^n-10^n}{3^{2n}} & = \lim_{n \to \infty} \dfrac{\dfrac{5^n}{3^{2n}}-\dfrac{10^n}{3^{2n}}}{\dfrac{3^{2n}}{3^{2n}}} \\ & = \lim_{n \to \infty} \dfrac{\left(\dfrac59\right)^n-\left(\dfrac{10}{9}\right)^n}{1} \\ & = \dfrac{0-\infty}{1} =-\infty \end{aligned}$
Thus, the value of $\boxed{\displaystyle \lim_{n \to \infty} \dfrac{5^n-10^n}{3^{2n}}=-\infty}$
Answer b)
Divide each term by $2^{2n}$, then apply the limit at infinity theorems.
$\begin{aligned} \displaystyle \lim_{n \to \infty} \dfrac{4+2^{2n}}{3^n-2} & = \lim_{n \to \infty} \dfrac{\dfrac{4}{2^{2n}}+\dfrac{2^{2n}}{2^{2n}}}{\dfrac{3^n}{2^{2n}}-\dfrac{2}{2^{2n}}} \\ & = \lim_{n \to \infty} \dfrac{\cancelto{0}{\dfrac{4}{4^n}} + \color{red}{1}}{\cancelto{0}{\left(\dfrac34\right)^n}-\cancelto{0}{\dfrac{2}{2^{2n}}}} \\ & = \infty \end{aligned}$
Note: The denominator in the last term tends to zero, making the limits tends to infinity, but we don’t suggest you to simply write $\dfrac{0+1}{0-0} = \infty$, as the answer is not “infinity”, but “undefined” since the denominator is exactly zero.

Thus, the value of $\boxed{\displaystyle \lim_{n \to \infty} \dfrac{4+2^{2n}}{3^n-2} = \infty}$
Answer c)
Divide each term by $(-3)^n$, then apply the limit at infinity theorems.
$\begin{aligned} \displaystyle \lim_{n \to \infty} \dfrac{(-3)^n}{2^n+1} & = \lim_{n \to \infty} \dfrac{\dfrac{(-3)^n}{(-3)^n}}{\dfrac{2^n}{(-3)^n}+\dfrac{1}{(-3)^n}} \\ & = \lim_{n \to \infty} \dfrac{1}{\left(-\dfrac23\right)^n + \dfrac{1}{(-3)^n}} \\ & = \pm \infty \end{aligned}$
The limit above does not exist, because for $n = 2k$ (even) and $n = 2k+1$ (odd), the limit is different in value.
Answer d)
Divide each term by $7^{n+1}$, then apply the limit at infinity theorems.
$\begin{aligned} & \displaystyle \lim_{n \to \infty} \dfrac{3^{n+1}+5^{n+1}+7^{n+1}}{3^n + 5^n + 7^n} \\ & = \lim_{n \to \infty} \dfrac{\dfrac{3^{n+1}}{7^{n+1}}+\dfrac{5^{n+1}}{7^{n+1}}+\dfrac{7^{n+1}}{7^{n+1}}}{\dfrac{3^n}{7^{n+1}} + \dfrac{5^n}{7^{n+1}} + \dfrac{7^n}{7^{n+1}}} \\ & = \dfrac{0+0+1}{0+0+\dfrac17} = 7 \end{aligned}$
Thus, the value of $\boxed{\displaystyle \lim_{n \to \infty} \dfrac{3^{n+1}+5^{n+1}+7^{n+1}}{3^n + 5^n + 7^n} = 7}$

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