Problem and Solution – Limit of Trigonometric Functions

      Here are the problems of limit involving trigonometric functions served along with the solutions to enhance the understanding of reader. The math formulas are generated by LaTeX system to smooth the way of writings.

Read: Problem & Solution – Limit at Infinity

Today Quote

Be the reason someday smiles today.

Problem Number 1
Evaluate $\displaystyle \lim_{x \to 0} \dfrac{\sin 3x \cdot \tan 5x} {1-\cos 5x}$.

Solution

Substitution method will fail since it leads to $\dfrac{0}{0}$ form. Hence, we need to manipute the form by recalling the following trigonometric identities and properties of limit involving trigonometric functions.
$\boxed{\begin{aligned} & \cos ax = 1-2 \sin^2 \dfrac{a}{2}x \\ & \displaystyle \lim_{x \to 0} \dfrac{\sin ax}{\sin bx} = \lim_{x \to 0} \dfrac{\tan ax}{\sin bx} = \dfrac{a}{b} \end{aligned}}$
We will have
$$\begin{aligned} \displaystyle \lim_{x \to 0} \dfrac{\sin 3x \cdot \tan 5x} {1-\cos 5x} & = \lim_{x \to 0} \dfrac{\sin 3x \cdot \tan 5x} {1-(1-2 \sin^2 \dfrac{5}{2}x)} \\ & = \lim_{x \to 0} \dfrac{\sin 3x \cdot \tan 5x}{2 \cdot \sin \dfrac{5}{2}x \cdot \sin \dfrac{5}{2}x} \\ & = \lim_{x \to 0} \left(\dfrac{1}{2} \cdot \dfrac{\sin 3x} {\sin \dfrac{5}{2}x} \cdot \dfrac{\tan 5x} {\sin \dfrac{5}{2}x}\right) \\ & = \dfrac{1}{2} \cdot \dfrac{6}{5} \cdot \dfrac{10}{5} = \dfrac{6}{5} \end{aligned}$$Thus, the value of $\boxed{\displaystyle \lim_{x \to 0} \dfrac{\sin 3x \cdot \tan 5x} {1-\cos 5x} = \dfrac{6}{5}} $

[collapse]

Problem Number 2
Evaluate $\displaystyle \lim_{x \to 0} \dfrac{1-\cos 2x} {x \tan 2x}$.

Solution

Substitution method will fail since it leads to $\dfrac{0}{0}$ form. Hence, we need to manipute the form by recalling the following trigonometric identities and properties of limit involving trigonometric functions.
$\boxed{\begin{aligned} \cos 2x & = 1-2 \sin^2 x \\ \displaystyle \lim_{x \to 0} \dfrac{\sin ax} {bx} & = \dfrac{a} {b} \\ \lim_{x \to 0} \dfrac{ax}{\tan bx} & = \dfrac{a}{b} \end{aligned}}$
We will have
$\begin{aligned} & \displaystyle \lim_{x \to 0} \dfrac{1-\cos 2x} {x \tan 2x} \\ & = \lim_{x \to 0} \dfrac{1-(1-2 \sin^2 x)} {x \tan 2x} \\ & = \lim_{x \to 0} \dfrac{2 \sin^2 x} {x \tan 2x} \\ & = \lim_{x \to 0} \left(2 \cdot \dfrac{\sin x} {x} \cdot \dfrac{\sin x}{\tan 2x}\right) \\ & = 2 \times 1 \times \dfrac{1}{2} = 1 \end{aligned}$
Thus, the value of $\boxed{\displaystyle \lim_{x \to 0} \dfrac{1-\cos 2x} {x \tan 2x} = 1}$

[collapse]

Problem Number 3
Evaluate $\displaystyle \lim_{x \to 3} \dfrac{x \tan (2x-6)} {\sin (x-3)}$.

Solution

Substitution method will fail since it leads to $\dfrac{0}{0}$ form. Hence, we need to manipute the form by recalling the following trigonometric identities and properties of limit involving trigonometric functions.
$\boxed{\displaystyle \lim_{x \to 0} \dfrac{\tan ax}{\sin bx} = \dfrac{a} {b}}$
We will have
$\begin{aligned} & \lim_{x \to 3} \dfrac{x \tan (2x-6)} {\sin (x-3)} \\ & = \lim_{x \to 3} \left(x \cdot \dfrac{\tan 2(x-3)} {\sin (x-3)}\right) \\ & = 3 \cdot 2 = 6 \end{aligned}$
Thus, the value of $\boxed{\displaystyle \lim_{x \to 3} \dfrac{x \tan (2x-6)} {\sin (x-3)} = 6}$

[collapse]

Problem Number 4
Evaluate $\displaystyle \lim_{x \to 0} \left(\dfrac{\sin 4x}{x^2 \tan 2x}- \dfrac{2}{x^2}\right)$.

Solution

Substitution method will fail since it leads to $\dfrac{0}{0}$ form. Hence, we need to manipute the form by recalling the following trigonometric identities and properties of limit involving trigonometric functions.
$\boxed{\begin{aligned} & \displaystyle \tan ax = \dfrac{\sin ax}{\cos ax} \\ & \sin 2ax = 2 \sin ax \cos ax \\ & \sin^2 ax + \cos^2 ax = 1 \\ & \lim_{x \to 0} \dfrac{\sin ax}{bx} = \dfrac{a}{b} \end{aligned}}$
We will have
$\begin{aligned} & \displaystyle \lim_{x \to 0} \left(\dfrac{\sin 4x}{x^2 \tan 2x}- \dfrac{2}{x^2}\right) \\ & = \lim_{x \to 0} \left(\dfrac{\sin 4x}{x^2 \tan 2x}- \dfrac{2 \tan 2x}{x^2 \tan 2x}\right) \\ & = \lim_{x \to 0} \dfrac{\sin 4x-2 \tan 2x}{x^2 \tan 2x} \\ & = \lim_{x \to 0} \dfrac{\sin 4x-2 \cdot \dfrac{\sin 2x} {\cos 2x}}{x^2 \tan 2x} \color{red}{\times \dfrac{\cos 2x}{\cos 2x}} \\ & = \lim_{x \to 0} \dfrac{\sin (2\cdot 2x) \cos 2x-2 \sin 2x}{x^2 \tan 2x \cos 2x} \\ & = \lim_{x \to 0} \dfrac{(2 \sin 2x \cos 2x) \cos 2x-2 sin 2x}{x^2 \tan 2x \cos 2x} \\ & = \lim_{x \to 0} \dfrac{2 \sin 2x(\cos^2 2x- 1)}{x^2 \tan 2x \cos 2x} \\ & = \lim_{x \to 0} \dfrac{2 \sin 2x (-\sin^2 2x)}{x^2 \cdot \dfrac{\sin 2x}{\cancel{\cos 2x}} \cancel{\cos 2x} } \\ & =-2 \lim_{x \to 0} \left(\dfrac{\sin 2x} {\sin 2x} \cdot \dfrac{\sin 2x} {x} \cdot \dfrac{\sin 2x} {x}\right) \\ & =-2 \cdot 1 \cdot 2 \cdot 2 =-8 \end{aligned}$
Thus, the value of $\boxed{\displaystyle \lim_{x \to 0} \left(\dfrac{\sin 4x}{x^2 \tan 2x}-\dfrac{2}{x^2}\right)=-8}$

[collapse]

Problem Number 5
Evaluate $\displaystyle \lim_{x \to 0} \dfrac{4x \cos x} {\sin x + \sin 3x}$.

Solution

Substitution method will fail since it leads to $\dfrac{0}{0}$ form. Hence, we need to manipute the form by recalling the following trigonometric identities and properties of limit involving trigonometric functions.
$$\boxed{\begin{aligned} & \sin ax + \sin bx = 2 \sin \left(\dfrac{a+b}{2}x\right) \cos \left(\dfrac{a-b}{2}x\right) \\ & \displaystyle \lim_{x \to 0} \dfrac{ax} {\sin bx} = \dfrac{a} {b} \end{aligned}}$$We will have
$$\begin{aligned} \displaystyle \lim_{x \to 0} \dfrac{4x \cos x} {\sin x + \sin 3x} & = \lim_{x \to 0} \dfrac{4x \cos x} {2 \sin \left(\dfrac{1+3}{2}x\right) \cos \left(\dfrac{1-3}{2}x\right)} \\ & = \lim_{x \to 0} \dfrac{4x \cdot \cancel{\cos x} } {2 \cdot \sin 2x \cdot \cancel{\cos x}} \\ & = \lim_{x \to 0} \left(\dfrac{4}{2} \cdot \dfrac{x} {\sin 2x} \right) \\ & = 2 \cdot \dfrac{1}{2} = 1 \end{aligned}$$Thus, the value of $\boxed{\displaystyle \lim_{x \to 0} \dfrac{4x \cos x} {\sin x + \sin 3x} = 1}$

[collapse]

Problem Number 6
Evaluate $\displaystyle \lim_{x \to 0} \dfrac{1- \cos 4x} {x \sin x}$.

Solution

Substitution method will fail since it leads to $\dfrac{0}{0}$ form. Hence, we need to manipute the form by recalling the following trigonometric identities and properties of limit involving trigonometric functions.
$\boxed{\begin{aligned} & \cos 2x = 1-2 \sin^2 x \\ & \displaystyle \lim_{x \to 0} \dfrac{\sin ax}{bx} = \dfrac{a}{b} \\ & \lim_{x \to 0} \dfrac{\sin ax}{\sin bx} = \dfrac{a}{b} \end{aligned}}$
We will have
$\begin{aligned} \displaystyle \lim_{x \to 0} \dfrac{1-\cos 4x}{x \sin x} & = \lim_{x \to 0} \dfrac{1-\cos (2 \cdot 2x)}{x \sin x} \\ & = \lim_{x \to 0} \dfrac{1-(1-2 \sin^2 2x)}{x \sin x} \\ & = \lim_{x \to 0} \dfrac{2 \sin^2 2x}{x \sin x} \\ & = \lim_{x \to 0} \left(2 \cdot \dfrac{\sin 2x}{x} \cdot \dfrac{\sin 2x}{\sin x}\right) \\ & = 2 \cdot 2 \cdot 2 = 8 \end{aligned}$
Thus, the value of $\boxed{\displaystyle \displaystyle \lim_{x \to 0} \dfrac{1-\cos 4x}{x \sin x}= 8}$

[collapse]

Problem Number 7
Evaluate $\displaystyle \lim_{x \to 0} \dfrac{x \cos 2x} {\tan x-\sin 2x}$.

Solution

Substitution method will fail since it leads to $\dfrac{0}{0}$ form. Hence, we need to manipute the form by recalling the following trigonometric identities and properties of limit involving trigonometric functions.
$\boxed{\begin{aligned} & \tan x = \dfrac{\sin x}{\cos x} \\ & \sin 2x = 2 \sin x \cos x \\ & \displaystyle \lim_{x \to 0} \dfrac{x}{\sin x} = 1 \\ & \lim_{x \to 0} \dfrac{\tan ax}{bx} = \lim_{x \to 0} \dfrac{\sin ax}{bx} = \dfrac{a}{b} \end{aligned}}$
Alternative I:

We will have
$$\begin{aligned} \displaystyle \lim_{x \to 0} \dfrac{x \cos 2x}{\tan x-\sin 2x} & = \lim_{x \to 0} \dfrac{x \cos 2x}{\dfrac{\sin x}{\cos x}-2 \sin x \cos x} \\ & = \lim_{x \to 0} \dfrac{x \cos 2x \cos x}{\sin x-2 \sin x \cos^2 x} \\ & = \lim_{x \to 0} \dfrac{x \cos 2x \cos x}{\sin x (1-2 \cos^2 x)} \\ & = \lim_{x \to 0} \left(\dfrac{x}{\sin x} \cdot \dfrac{\cos 2x \cos x}{1-2 \cos^2 x}\right) \\ & = 1 \cdot \dfrac{\cos 0 \cdot \cos 0}{1-2 \cos^2 0} \\ & = 1 \cdot \dfrac{1 \cdot 1}{1-2 \cdot 1} =-1 \end{aligned}$$Alternative II:
$$\begin{aligned} \displaystyle \lim_{x \to 0} \dfrac{x \cos 2x} {\tan x-\sin 2x} & = \lim_{x \to 0} \left(\dfrac{x \cos 2x} {\tan x-\sin 2x} \cdot \dfrac{\frac{1}{x}} {\frac{1}{x}}\right) \\ & = \lim_{x \to 0} \dfrac{\cos 2x} {\frac{\tan x} {x}-\frac{\sin 2x} {x}} \\ & = \lim_{x \to 0} \dfrac{\cos 0}{1-2} = \dfrac{1}{-1} =-1 \end{aligned}$$Thus, the value of $\boxed{\displaystyle \lim_{x \to 0} \dfrac{x \cos 2x}{\tan x-\sin 2x} =-1}$

[collapse]

Problem Number 8
Evaluate $\displaystyle \lim_{x \to 1} \dfrac{(x^2-1) \tan (2x-2)} {\sin^2 (x-1)}$.

Solution

Substitution method will fail since it leads to $\dfrac{0}{0}$ form. Hence, we need to manipute the form by recalling the following properties of limit involving trigonometric functions.
$\boxed{\begin{aligned} & \displaystyle \lim_{x \to 0} \dfrac{x}{\sin x} =1 \\ & \lim_{x \to 0} \dfrac{\tan ax}{\sin bx} = \dfrac{a}{b} \end{aligned}}$
We will have
$$\begin{aligned} & \displaystyle \lim_{x \to 1} \dfrac{(x^2-1) \tan (2x-2)}{\sin^2 (x-1)} \\ & = \lim_{x \to 1} \dfrac{(x+1)(x-1) \tan 2(x-1)}{\sin (x-1) \cdot \sin (x-1)} \\ & = \lim_{x \to 1} \left((x + 1) \cdot \dfrac{x-1}{\sin (x-1)} \cdot \dfrac{\tan 2(x-1)}{\sin (x-1)}\right) \\ & = (1 + 1) \cdot 1 \cdot 2 = 4 \end{aligned}$$Thus, the value of $\boxed{\displaystyle \lim_{x \to 1} \dfrac{(x^2-1) \tan(2x-2)}{\sin^2 (x-1)} = 4}$

[collapse]

Problem Number 9
Evaluate $\displaystyle \lim_{x \to \frac{\pi} {4}} \dfrac{\cos 2x} {\sin x- \cos x}$.

Solution

Substitution method will fail since it leads to $\dfrac{0}{0}$ form. Hence, we need to manipute the form by recalling the following trigonometric identities.
$\boxed{\cos 2x = \cos^2 x-\sin^2 x}$
By multiplying the function to its denominator’s conjugate, we have
$$\begin{aligned} \displaystyle \lim_{x \to \frac{\pi} {4}} \dfrac{\cos 2x} {\sin x-\cos x} & = \lim_{x \to \frac{\pi} {4}} \left(\dfrac{\cos 2x} {\sin x-\cos x} \times \dfrac{\sin x + \cos x} {\sin x + \cos x} \right) \\ & = \lim_{x \to \frac{\pi} {4}} \dfrac{\cos 2x (\sin x + \cos x)} {\sin^2 x- \cos^2 x} \\ & = \lim_{x \to \frac{\pi} {4}} \dfrac{\cancel{\cos 2x} (\sin x + \cos x)} {-\cancel{\cos 2x}} \\ & = \lim_{x \to \frac{\pi} {4}}-(\sin x+ \cos x) \\ & =-\left(\sin \dfrac{\pi} {4} + \cos \dfrac{\pi} {4}\right) \\ & =-\left(\dfrac{1}{2}\sqrt{2} + \dfrac{1}{2}\sqrt{2}\right) =-\sqrt{2} \end{aligned}$$Thus, the value of $\boxed{\displaystyle \lim_{x \to \frac{\pi} {4}} \dfrac{\cos 2x} {\sin x- \cos x} =-\sqrt{2}}$

[collapse]

Problem Number 10
Evaluate $\displaystyle \lim_{x \to \frac{\pi} {8}} \dfrac{\sin^2 2x-\cos^2 2x} {\sin 2x-\cos 2x}$.

Solution

Substitution method will fail since it leads to $\dfrac{0}{0}$ form. 
By multiplying the function to its denominator’s conjugate, we have
$$\begin{aligned} \displaystyle & \lim_{x \to \frac{\pi} {8}} \dfrac{\sin^2 2x-\cos^2 2x} {\sin 2x-\cos 2x} \\ & = \lim_{x \to \frac{\pi} {8}} \left(\dfrac{\sin^2 2x-\cos^2 2x} {\sin 2x- \cos 2x} \times \dfrac{\sin 2x + \cos 2x}{\sin 2x + \cos 2x}\right) \\ & = \lim_{x \to \frac{\pi} {8}} \dfrac{\cancel{(\sin^2 2x-\cos^2 2x)}(\sin 2x + \cos 2x)}{\cancel{\sin^2 2x-\cos^2 2x}} \\ & = \lim_{x \to \frac{\pi} {8}} (\sin 2x + \cos 2x) \\ & = \sin \dfrac{2\pi}{8} + \cos \dfrac{2\pi}{8} \\ & = \dfrac{1}{2}\sqrt{2} + \dfrac{1}{2}\sqrt{2} = \sqrt{2} \end{aligned}$$Thus, the value of $\boxed{\displaystyle \lim_{x \to \frac{\pi} {8}} \dfrac{\sin^2 2x- \cos^2 2x} {\sin 2x-\cos 2x} = \sqrt{2}}$

[collapse]

Problem Number 11
Evaluate $\displaystyle \lim_{x \to 2} \dfrac{(x-2) \cos (\pi x-2\pi)} {\tan (2\pi x- 4\pi)}$.

Solution

Substitution method will fail since it leads to $\dfrac{0}{0}$ form. Hence, we need to manipute the form by recalling the following properties of limit involving trigonometric functions.
$\boxed{\displaystyle \lim_{x \to 0} \dfrac{ax}{\tan bx} = \dfrac{a}{b}}$
We will have
$\begin{aligned} & \displaystyle \lim_{x \to 2} \dfrac{(x-2) \cos (\pi x-2\pi)} {\tan (2\pi x- 4\pi)} \\ & = \lim_{x \to 2} \dfrac{(x-2) \cos \pi(x-2)}{\tan 2\pi(x-2)} \\ & = \lim_{x \to 2} \left(\dfrac{x-2}{\tan 2\pi(x-2)} \cdot \cos \pi(x-2)\right) \\ & = \dfrac{1}{2\pi} \cdot 1 = \dfrac{1}{2\pi} \end{aligned}$
Thus, the value of $\boxed{\displaystyle \lim_{x \to 2} \dfrac{(x-2) \cos (\pi x-2\pi)} {\tan (2\pi x-4\pi)} = \dfrac{1}{2\pi}}$

[collapse]

Problem Number 12
Evaluate $\displaystyle \lim_{x \to 0} \dfrac{4x \cos x} {\sin x + \sin 3x}$.

Solution

Substitution method will fail since it leads to $\dfrac{0}{0}$ form. Hence, we need to manipute the form by recalling the following trigonometric identities and properties of limit involving trigonometric functions.
$$\boxed{\begin{aligned} & \sin ax + \sin bx = 2 \sin \left(\dfrac{a+b}{2}x\right) \cos \left(\dfrac{a-b}{2}x\right) \\ & \cos (-x) = \cos x \\ & \displaystyle \lim_{x \to 0} \dfrac{ax} {\sin bx} = \dfrac{a} {b} \end{aligned}}$$We will have
$\begin{aligned} & \displaystyle \lim_{x \to 0} \dfrac{4x \cos x} {\sin x + \sin 3x} \\ & = \lim_{x \to 0} \dfrac{4x \cdot \cos x} {2 \sin \left(\dfrac{1+3}{2}x\right) \cos \left(\dfrac{1-3}{2}x\right)}\ \\ & = \lim_{x \to 0} \dfrac{4x \cdot \cancel{\cos x} } {2 \sin 2x \cancel{\cos x}} \\ & = \lim_{x \to 0} \left(\dfrac{4}{2} \cdot \dfrac{x} {\sin 2x} \right) \\ & = 2 \cdot \dfrac{1}{2} = 1 \end{aligned}$
Thus, the value of $\boxed{\displaystyle \lim_{x \to 0} \dfrac{4x \cos x} {\sin x + \sin 3x} = 1}$

[collapse]

Problem Number 13
Evaluate $\displaystyle \lim_{x \to 0} \dfrac{3x + \sin 4x}{5x-\tan 2x}$.

Solution

Divide every term in the numerator and denominator by $x$, so that we are able to apply the properties of limit involving trigonometric functions.
$\begin{aligned} \displaystyle \lim_{x \to 0} \dfrac{3x + \sin 4x}{5x-\tan 2x} & = \lim_{x \to 0} \dfrac{3 + \dfrac{\sin 4x}{x}}{5- \dfrac{2 \tan 2x}{2x}} \\ & = \lim_{x \to 0} \dfrac{3 + \dfrac{4 \sin 4x}{4x}}{5- \dfrac{\tan 2x}{x}} \\ & = \dfrac{3 + 4 \cdot \displaystyle \lim_{4x \to 0} \dfrac{\sin 4x}{4x}}{5-2 \cdot \displaystyle \lim_{2x \to 0} \dfrac{\tan 2x}{2x}} \\ & = \dfrac{3+4}{5-2} = \dfrac{7}{3} \end{aligned}$
Thus, the value of $\boxed{\displaystyle \lim_{x \to 0} \dfrac{3x + \sin 4x}{5x-\tan 2x} = \dfrac{7}{3}}$

[collapse]

Problem Number 14
Evaluate $\displaystyle \lim_{x \to \frac{\pi}{2}} \dfrac{\cos x}{x-\dfrac{\pi}{2}}$.

Solution

Substitution method will fail since it leads to $\dfrac{0}{0}$ form. Hence, we need to manipute the form by recalling the following trigonometric identities.
$\boxed{\cos x = \sin \left(\dfrac{\pi}{2}- x\right)}$
We will have
$\begin{aligned} \displaystyle \lim_{x \to \frac{\pi}{2}} \dfrac{\cos x}{x-\dfrac{\pi}{2}} & = \lim_{x \to \frac{\pi}{2}} \dfrac{\sin \left(\dfrac{\pi}{2}-x\right)}{x-\dfrac{\pi}{2}} \\ & = \lim_{x \to \frac{\pi}{2}} \dfrac{\sin \left(\dfrac{\pi}{2}-x\right)}{-\left(\dfrac{\pi}{2}-x\right)} \\ & =-\lim_{x \to \frac{\pi}{2}} \dfrac{\sin \left(\dfrac{\pi}{2}-x\right)}{\left(\dfrac{\pi}{2}-x\right)} =-1 \end{aligned}$
Thus, the value of $\boxed{\displaystyle \lim_{x \to \frac{\pi}{2}} \dfrac{\cos x}{x- \dfrac{\pi}{2}} =-1}$

[collapse]

Problem Number 15
Evaluate $\displaystyle \lim_{a \to 0} \dfrac{1}{a} \left(\dfrac{\sin^3 2a} {\cos 2a} + \sin 2a \cos 2a\right)$.

Solution

Recall that $\boxed{\displaystyle \lim_{x \to 0} \dfrac{\sin ax}{bx} = \dfrac{a} {b}}$. 
Hence, we have
$\begin{aligned} & \displaystyle \lim_{a \to 0} \dfrac{1}{a} \left(\dfrac{\sin^3 2a} {\cos 2a} + \sin 2a \cos 2a\right) \\ & = \lim_{a \to 0} \dfrac{\sin 2a} {a} \left(\dfrac{\sin^2 2a} {\cos 2a} + \cos 2a\right) \\ & = \lim_{a \to 0} \dfrac{\sin 2a} {a} \cdot \lim_{a \to 0} \left(\dfrac{\sin^2 2a} {\cos 2a} + \cos 2a\right) \\ & = 2(0+1) = 2 \end{aligned}$
Thus, the value of $\boxed{\displaystyle \lim_{a \to 0} \dfrac{1}{a} \left(\dfrac{\sin^3 2a} {\cos 2a} + \sin 2a \cos 2a\right) = 2}$

[collapse]

Problem Number 16
Tentukan nilai dari $\displaystyle \lim_{x \to 0} \dfrac{x^2 \tan 2x} {x-x \cos 4x}$.

Solution

Substitution method will fail since it leads to $\dfrac{0}{0}$ form. Hence, we need to manipute the form by recalling the following trigonometric identities and properties of limit involving trigonometric functions.
$\boxed{\begin{aligned} \cos ax & = 1-2 \sin^2 \dfrac{a}{2}x \\ \lim_{x \to 0} \dfrac{ax}{\sin bx} & = \lim_{x \to 0} \dfrac{\tan ax}{\sin bx} = \dfrac{a}{b} \end{aligned}}$
We will have
$$\begin{aligned} \displaystyle \lim_{x \to 0} \dfrac{x^2 \tan 2x} {x-x \cos 4x} & = \lim_{x \to 0} \dfrac{\cancel{x} \cdot x \tan 2x} {\cancel{x} (1-\cos 4x)} \\ & = \lim_{x \to 0} \dfrac{x \tan 2x} {1-(1-2 \sin^2 2x)} \\ & \lim_{x \to 0} \dfrac{x \tan 2x}{2 \sin^2 2x} \\ & = \lim_{x \to 0} \left(\dfrac{1}{2} \cdot \dfrac{x} {\sin 2x} \cdot \dfrac{\tan 2x} {\sin 2x} \right) \\ & = \dfrac{1}{2} \cdot \dfrac{1}{2} \cdot 1 = \dfrac{1}{4} \end{aligned}$$Thus, the value of $\displaystyle \lim_{x \to 0} \dfrac{x^2 \tan 2x} {x-x \cos 4x}$ adalah $\dfrac{1}{4}$

[collapse]
 

Problem Number 17
Evaluate $\displaystyle \lim_{x \to 45^{\circ}} \dfrac{\cos 2x} {1-\tan x}$.

Solution

Substitution method will fail since it leads to $\dfrac{0}{0}$ form. Hence, we need to manipute the form by recalling the following trigonometric identities.
$\boxed{\begin{aligned} \tan x & = \dfrac{\sin x} {\cos x} \\ \cos 2x & = \cos^2 x- \sin^2 x \end{aligned}}$
We will have
$\begin{aligned} & \displaystyle \lim_{x \to 45^{\circ}} \dfrac{\cos 2x} {1-\tan x} \\ & = \lim_{x \to 45^{\circ}} \dfrac{\cos^2 x- \sin^2 x}{\dfrac{\cos x-\sin x} {\cos x}} \\ & = \lim_{x \to 45^{\circ}} \dfrac{(\cos^2 x- \sin^2 x) \cos x} {\cos x-\sin x} \\ & = \lim_{x \to 45^{\circ}} \dfrac{(\cos x + \sin x) \cancel{(\cos x-\sin x)} \cos x} {\cancel{\cos x-\sin x}} \\ & = \lim_{x \to 45^{\circ}} (\cos x + \sin x) \cos x \\ & = (\cos 45^{\circ} + \sin 45^{\circ}) \cos 45^{\circ} \\ & = \left(\dfrac{1}{2}\sqrt{2} + \dfrac{1}{2}\sqrt{2}\right) \cdot \dfrac{1}{2}\sqrt{2} = 1 \end{aligned}$
Thus, the value of $\displaystyle \lim_{x \to 45^{\circ}} \dfrac{\cos 2x} {1-\tan x}$ adalah $\boxed{1}$

[collapse]

Problem Number 18
The value of $\displaystyle \lim_{x \to 1} \dfrac{\tan (1-x)}{x^3-1} = \cdots \cdot$
A. $\dfrac{1}{3}$                    C. $1$                  E. $\dfrac{1}{2}$
B. $-\dfrac{1}{3}$                 D. $-1$    

Solution

Note that $x^3- 1$ can be factorized into $(x-1)(x^2 + x + 1)$, so by using factorize method in evaluating the limit value, we have
$\begin{aligned} & \displaystyle \lim_{x \to 1} \dfrac{\tan (1-x)}{x^3-1} \\ & = \lim_{x \to 1} \dfrac{\tan (1-x)}{(x-1)(x^2+x+1)} \\ & = \lim_{x \to 1} \left(-\dfrac{\tan (x-1)}{x-1} \cdot \dfrac{1}{x^2 + x + 1}\right) \\ & =-\dfrac{1}{1^2 + 1 + 1} =-\dfrac{1}{3} \end{aligned}$
Thus, the value of $\boxed{\displaystyle \lim_{x \to 1} \dfrac{\tan (1-x)}{x^3-1} =-\dfrac{1}{3}}$
Trivia: Also recall that $\boxed{\lim_{x \to 0} \dfrac{\tan x}{x} = 1}$
(Answer C)

[collapse]

Problem Number 19
The value $\displaystyle \lim_{x \to 0} \dfrac{\sec 9x-\sec 7x}{\sec 5x-\sec 3x}$ is $\cdots \cdot$
A. $0$                    C. $2$                    E. $4$
B. $1$                    D. $3$           

Solution

Substitution method will fail since it leads to $\dfrac{0}{0}$ form. Hence, we need to manipute the form by recalling the following trigonometric identities and the sum of cosines functions.
$$\boxed{\begin{aligned} \sec x & = \dfrac{1}{\cos x} \\ \cos x + \cos y & = 2 \cos \dfrac12(x+y) \cos \dfrac12(x-y) \end{aligned}}$$ $$\begin{aligned} & \displaystyle \lim_{x \to 0} \dfrac{\sec 9x-\sec 7x}{\sec 5x-\sec 3x} \\ & = \lim_{x \to 0} \dfrac{\dfrac{1}{\cos 9x}-\dfrac{1}{\cos 7x}}{\dfrac{1}{\cos 5x}-\dfrac{1}{\cos 3x}} \\ & = \lim_{x \to 0} \dfrac{\dfrac{\cos 7x-\cos 9x}{\cos 9x \cos 7x}}{\dfrac{\cos 3x-\cos 5x}{\cos 3x \cos 5x}} \\ & = \lim_{x \to 0} \dfrac{\cos 7x- \cos 9x}{\cos 9x \cos 7x} \times \dfrac{\cos 3x \cos 5x}{\cos 3x- \cos 5x} \\ & = \lim_{x \to 0} \dfrac{-2 \sin \dfrac12(7x+9x) \sin \dfrac12(7x-9x)}{\cos 9x \cos 7x} \times \dfrac{\cos 3x \cos 5x}{-2 \sin \dfrac12(3x+5x) \sin \dfrac12(3x-5x)} \\ & = \lim_{x \to 0} \dfrac{-2 \sin 8x \sin (-x)}{\cos 9x \cos 7x} \times \dfrac{\cos 3x \cos 5x}{-2 \sin 4x \sin (-x)} \\ & = \lim_{x \to 0} \dfrac{\sin 8x}{\sin 4x} \times \dfrac{\sin x}{\sin x} \times \dfrac{\cos 3x \cos 5x}{\cos 9x \cos 7x} \\ & = \dfrac{8}{4} \times 1 \times \dfrac{\cos 0 \cos 0}{\cos 0 \cos 0} \\ & = 2 \end{aligned}$$Thus, the value of $\boxed{\displaystyle \lim_{x \to 0} \dfrac{\sec 9x-\sec 7x}{\sec 5x-\sec 3x} = 2}$
(Answer C)

[collapse]

Problem Number 20
Evaluate the following limits.
a. $\displaystyle \lim_{x \to 0} \dfrac{\sin(4x^2)}{x^2+\tan^2(3x)}$
b. $\displaystyle \lim_{x \to 0} \dfrac{\tan(2x^3)}{x^3+\sin^3(4x)}$

Solution

Answer a)
Divide both numerator and denominator by $x^2$ and note that

$\begin{aligned} \displaystyle \lim_{x \to 0} \dfrac{\sin ax^2}{bx^2} & = \dfrac{a}{b} \\ \lim_{x \to 0} \dfrac{\tan ax}{bx} & = \dfrac{a}{b} \end{aligned}$
By applying these, we will have
$$\begin{aligned} \displaystyle \lim_{x \to 0} \dfrac{\dfrac{\sin(4x^2)}{x^2}}{\dfrac{x^2}{x^2}+\dfrac{\tan^2(3x)}{x^2}} & = \displaystyle \lim_{x \to 0} \dfrac{\dfrac{\sin(4x^2)}{x^2}}{\dfrac{x^2}{x^2}+\dfrac{\tan 3x}{x} \times \dfrac{\tan 3x}{x}} \\ & = \dfrac{4}{1 + 3 \times 3} = \dfrac{4}{10} = \dfrac25 \end{aligned}$$Thus, the value of $\boxed{\displaystyle \lim_{x \to 0} \dfrac{\sin(4x^2)}{x^2+\tan^2(3x)}= \dfrac25}$
Answer b)
Divide both numerator and denominator by $x^2$ and note that
$\begin{aligned} \displaystyle \lim_{x \to 0} \dfrac{\tan ax^3}{bx^3} & = \dfrac{a}{b} \\ \lim_{x \to 0} \dfrac{\sin ax}{bx} & = \dfrac{a}{b} \end{aligned}$
By applying these, we will have
$$\begin{aligned} \displaystyle \lim_{x \to 0} \dfrac{\tan(2x^3)}{x^3+\sin^3(4x)} & = \lim_{x \to 0} \dfrac{\dfrac{\tan (2x^3)}{x^3}}{\dfrac{x^3}{x^3} + \dfrac{\sin^3 (4x)}{x^3}} \\ & = \lim_{x \to 0} \dfrac{\dfrac{\tan (2x^3)}{x^3}}{\dfrac{x^3}{x^3} + \dfrac{\sin (4x)}{x} \times \dfrac{\sin (4x)}{x} \times \dfrac{\sin (4x)}{x}} \\ & = \dfrac{2}{1 + 4 \times 4 \times 4} = \dfrac{2}{65} \end{aligned}$$Thus, the value of $\boxed{\displaystyle \lim_{x \to 0} \dfrac{\tan(2x^3)}{x^3+\sin^3(4x)}=\dfrac{2}{65}}$

[collapse]

Problem Number 21
The value of $\displaystyle \lim_{x \to \frac{\pi}{2}} (\pi-2x) \cdot \tan 5x = \cdots \cdot$
A. $\dfrac45$                   C. $\dfrac25$                  E. $\dfrac14$
B. $\dfrac35$                   D. $\dfrac12$         

Solution

Suppose $u = x-\dfrac{\pi}{2}$, or in the other words $x = u + \dfrac{\pi}{2}$. When $x$ approaches $\dfrac{\pi}{2}$, $u$ will surely approach $0$. Hence, the limit function can be evaluated.
$$\begin{aligned} & \displaystyle \lim_{u \to 0} \left(\pi-2\left(u + \dfrac{\pi}{2}\right)\right) \tan 5\left(u+\dfrac{\pi}{2}\right) \\ & =-2 \lim_{u \to 0} u \tan \left(5u + \dfrac{\pi}{2}\right) && \left(\cdots \tan \dfrac{5\pi}{2} = \tan \dfrac{\pi}{2}\right) \\ & =-2 \lim_{u \to 0} u (-\cot 5u) && \left(\cdots \tan \left(u + \dfrac{\pi}{2}\right) =-\cot u\right) \\ & = 2 \lim_{u \to 0} \dfrac{u}{\tan 5u} \\ & = 2 \times \dfrac15 = \dfrac25 \end{aligned}$$Thus, the value of $\boxed{\displaystyle \lim_{x \to \frac{\pi}{2}} (\pi-2x) \cdot \tan 5x = \dfrac25}$
(Answer C)

[collapse]

CategoriesLimit Fungsi, TrigonometriTags, ,

Leave a Reply

Silakan beri tanggapan dan saran, tidak perlu sungkan. Mohon juga diinformasikan melalui kolom komentar ini bila ada kesalahan pengetikan sekecil apapun (typo atau bahasa latex yang error) atau kesalahan konsep dan pembahasan soal. Terima kasih. Ganbatte!

Your email address will not be published. Required fields are marked *