Problem and Solution – Limit of Trigonometric Functions

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}} $

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 $\boxed{\displaystyle \lim_{x \to 0} \dfrac{1-\cos 2x} {x \tan 2x} = 1}$

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 $\boxed{\displaystyle \lim_{x \to 3} \dfrac{x \tan (2x-6)} {\sin (x-3)} = 6}$

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 $\boxed{\displaystyle \lim_{x \to 0} \left(\dfrac{\sin 4x}{x^2 \tan 2x}-\dfrac{2}{x^2}\right)=-8}$

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 $\boxed{\displaystyle \lim_{x \to 0} \dfrac{4x \cos x} {\sin x + \sin 3x} = 1}$

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 $\boxed{\displaystyle \displaystyle \lim_{x \to 0} \dfrac{1-\cos 4x}{x \sin x}= 8}$

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 $\boxed{\displaystyle \lim_{x \to 0} \dfrac{x \cos 2x}{\tan x-\sin 2x} =-1}$

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 $\boxed{\displaystyle \lim_{x \to 1} \dfrac{(x^2-1) \tan(2x-2)}{\sin^2 (x-1)} = 4}$

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 $\boxed{\displaystyle \lim_{x \to \frac{\pi} {4}} \dfrac{\cos 2x} {\sin x- \cos x} =-\sqrt{2}}$

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 $\boxed{\displaystyle \lim_{x \to \frac{\pi} {8}} \dfrac{\sin^2 2x- \cos^2 2x} {\sin 2x-\cos 2x} = \sqrt{2}}$

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 $\boxed{\displaystyle \lim_{x \to 2} \dfrac{(x-2) \cos (\pi x-2\pi)} {\tan (2\pi x-4\pi)} = \dfrac{1}{2\pi}}$

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 $\boxed{\displaystyle \lim_{x \to 0} \dfrac{4x \cos x} {\sin x + \sin 3x} = 1}$

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 $\boxed{\displaystyle \lim_{x \to 0} \dfrac{3x + \sin 4x}{5x-\tan 2x} = \dfrac{7}{3}}$

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 $\boxed{\displaystyle \lim_{x \to \frac{\pi}{2}} \dfrac{\cos x}{x- \dfrac{\pi}{2}} =-1}$

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 $\boxed{\displaystyle \lim_{a \to 0} \dfrac{1}{a} \left(\dfrac{\sin^3 2a} {\cos 2a} + \sin 2a \cos 2a\right) = 2}$

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 $\dfrac{1}{4}$

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 $\boxed{1}$

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)

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)

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 $\boxed{\displaystyle \lim_{x \to 0} \dfrac{\tan(2x^3)}{x^3+\sin^3(4x)}=\dfrac{2}{65}}$

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)