Category: Limit Fungsi

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)

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)

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

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)

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)

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)

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)

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)

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)

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)

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)

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)

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

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)

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)

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)

$\begin{aligned} & \displaystyle \lim_{x \to \infty} \dfrac{3^{x+1} + 2^x-3}{3^{x+2}-2^{x-1} + 4} \\ & = \lim_{x \to \infty} \dfrac{3^{x+1} + 2^x-3}{3^{x+2}-2^{x-1} + 4} \times \dfrac{\frac{1}{3^{x+2}}} {\frac{1}{3^{x+2}}} \\ & = \lim_{x \to \infty} \dfrac{\frac13 + \frac{2^x} {3^{x+2}}-\dfrac{3}{3^{x+2}}} {1-\frac{2^{x-1}}{3^{x+2}} + \dfrac{4}{3^{x+2}}} \\ & = \dfrac{\frac13 + 0-0}{1-0 + 0} = \dfrac13 \end{aligned}$
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)

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)

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)

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)

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)

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)

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)

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)

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)

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)

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)

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)

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

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)

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)

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) 
 

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)

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)

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)

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

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
$\displaystyle \lim_{x \to \infty} \dfrac{x^5-2x^4+x^3-3x^2+2x-7}{7-2x+3x^2-x^3+2x^4} = \infty$

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) 
$\displaystyle \lim_{x \to \infty} \dfrac{(3x-2)^3}{(4x+2)^3} = \lim_{x \to \infty} \dfrac{27x^3 + \cdots} {64x^3 + \cdots} = \dfrac{27}{64}$

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,
$\displaystyle \lim_{x \to \infty} (\sqrt{x+5}-\sqrt{2x-3}) =-\infty$

$$\begin{aligned} \displaystyle \lim_{\theta \to-\infty} \dfrac{\pi \theta^5}{\theta^5-5\theta^4} & = \lim_{\theta \to-\infty} \dfrac{\cancel{\theta^4}(\pi \theta)}{\cancel{\theta^4}(\theta-5)} \\ & = \lim_{\theta \to-\infty} \dfrac{\pi \theta} {\theta-5} \\ & = \lim_{\theta \to-\infty} \left(\dfrac{\pi(\theta-5)} {\theta-5} + \dfrac{5\pi} {\theta-5}\right) \\ & = \lim_{\theta \to-\infty} \left(\pi + \dfrac{5\pi} {\theta-5}\right) \\ & = \pi-0 = \pi \end{aligned}$$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 $0$.

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 $\boxed{\displaystyle \lim_{x \to \infty} (x-\sqrt{x^2-10x }) = 5}$

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 $\boxed{\displaystyle \lim_{x \to \infty} \dfrac{\sqrt{8x^2+1}} {x^2+4} = 0}$

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

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 $\boxed{\sqrt{3}}$

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 $\boxed{\dfrac{2}{7}}$

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 $\boxed{\displaystyle \lim_{x \to \infty} \dfrac{\csc \frac{1}{x}} {x} = 1}$  

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 $\boxed{\displaystyle \lim_{x \to \infty} \dfrac{\cot \dfrac{1}{2x}} {\csc \dfrac{3}{x}} = 6}$