Limit of trigonometric function Questions in English

(C) We evaluate the limit: $\mathop {\lim }\limits_{x \to 0} \frac{{{x^3}\cot x}}{{1 - \cos x}}$.
Using the identity $1 - \cos x = 2\sin^2(x/2)$ or multiplying by the conjugate:
$\mathop {\lim }\limits_{x \to 0} \frac{{{x^3}\cos x}}{{\sin x (1 - \cos x)}} = \mathop {\lim }\limits_{x \to 0} \left( \frac{x}{\sin x} \right) \cdot \frac{x^2}{1 - \cos x} \cdot \cos x$.
Since $\mathop {\lim }\limits_{x \to 0} \frac{x}{\sin x} = 1$ and $\mathop {\lim }\limits_{x \to 0} \frac{x^2}{1 - \cos x} = \mathop {\lim }\limits_{x \to 0} \frac{x^2}{2\sin^2(x/2)} = \mathop {\lim }\limits_{x \to 0} \frac{2}{( \sin(x/2) / (x/2) )^2} = 2$.
Thus,the limit is $1 \times 2 \times \cos(0) = 1 \times 2 \times 1 = 2$.

(A) We know that $1 - \cos 2x = 2 \sin^2 x$.
So,the limit becomes $\mathop {\lim }\limits_{x \to 0} \frac{2 \sin^2 x}{x}$.
This can be rewritten as $\mathop {\lim }\limits_{x \to 0} \left( 2 \cdot \frac{\sin x}{x} \cdot \sin x \right)$.
Using the standard limit $\mathop {\lim }\limits_{x \to 0} \frac{\sin x}{x} = 1$,we get $2 \cdot 1 \cdot \sin(0) = 2 \cdot 1 \cdot 0 = 0$.

(C) Given: $\mathop {\lim }\limits_{x \to 0} kx\,\text{cosec}\,x = \mathop {\lim }\limits_{x \to 0} x\,\text{cosec}\,kx$
We know that $\text{cosec}\,\theta = \frac{1}{\sin \theta}$.
So,$\mathop {\lim }\limits_{x \to 0} \frac{kx}{\sin x} = \mathop {\lim }\limits_{x \to 0} \frac{x}{\sin kx}$
Multiply and divide the right side by $k$: $\mathop {\lim }\limits_{x \to 0} \frac{kx}{\sin x} = \frac{1}{k} \mathop {\lim }\limits_{x \to 0} \frac{kx}{\sin kx}$
Using the standard limit $\mathop {\lim }\limits_{\theta \to 0} \frac{\theta}{\sin \theta} = 1$,we get:
$k(1) = \frac{1}{k}(1)$
$k = \frac{1}{k}$
$k^2 = 1$
$k = \pm 1$

(D) We know that $\mathop {\lim }\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$.
To evaluate $\mathop {\lim }\limits_{x \to 0} \frac{\sin 2x}{x}$,we multiply and divide by $2$:
$\mathop {\lim }\limits_{x \to 0} \frac{\sin 2x}{x} = \mathop {\lim }\limits_{x \to 0} \left( 2 \times \frac{\sin 2x}{2x} \right)$.
Since as $x \to 0$,$2x \to 0$,we have:
$2 \times \mathop {\lim }\limits_{2x \to 0} \frac{\sin 2x}{2x} = 2 \times 1 = 2$.

(C) Using the trigonometric identity $1 - \cos \theta = 2 \sin^2(\theta/2)$:
$\mathop {\lim }\limits_{x \to 0} \frac{1 - \cos mx}{1 - \cos nx} = \mathop {\lim }\limits_{x \to 0} \frac{2 \sin^2(mx/2)}{2 \sin^2(nx/2)}$
$= \mathop {\lim }\limits_{x \to 0} \left[ \left( \frac{\sin(mx/2)}{mx/2} \right)^2 \cdot \frac{m^2 x^2}{4} \cdot \left( \frac{nx/2}{\sin(nx/2)} \right)^2 \cdot \frac{4}{n^2 x^2} \right]$
Since $\mathop {\lim }\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$,we get:
$= \frac{m^2}{n^2} \cdot 1 = \frac{m^2}{n^2}$.
Alternative method: Apply $L$-Hospital's rule:
$\mathop {\lim }\limits_{x \to 0} \frac{1 - \cos mx}{1 - \cos nx} = \mathop {\lim }\limits_{x \to 0} \frac{m \sin mx}{n \sin nx}$
$= \mathop {\lim }\limits_{x \to 0} \frac{m^2 \cos mx}{n^2 \cos nx} = \frac{m^2(1)}{n^2(1)} = \frac{m^2}{n^2}$.

(C) We know that $\mathop {\lim }\limits_{\theta \to 0} \frac{{\sin \theta }}{\theta } = 1$.
Given expression is $\mathop {\lim }\limits_{x \to 0} \frac{{2{{\sin }^2}3x}}{{{x^2}}}$.
$= 2 \times \mathop {\lim }\limits_{x \to 0} \left( \frac{{\sin 3x}}{x} \right)^2$.
$= 2 \times \mathop {\lim }\limits_{x \to 0} \left( 3 \times \frac{{\sin 3x}}{{3x}} \right)^2$.
$= 2 \times 3^2 \times \left( \mathop {\lim }\limits_{3x \to 0} \frac{{\sin 3x}}{{3x}} \right)^2$.
$= 2 \times 9 \times (1)^2 = 18$.

(A) We know that $\mathop {\lim }\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$.
Given limit is $\mathop {\lim }\limits_{x \to 0} \frac{\sin ax}{\sin bx}$.
Multiply and divide by $ax$ and $bx$:
$= \mathop {\lim }\limits_{x \to 0} \left( \frac{\sin ax}{ax} \cdot ax \cdot \frac{1}{\frac{\sin bx}{bx} \cdot bx} \right)$
$= \mathop {\lim }\limits_{x \to 0} \left( \frac{\sin ax}{ax} \right) \cdot \left( \frac{bx}{\sin bx} \right) \cdot \frac{ax}{bx}$
$= 1 \cdot 1 \cdot \frac{a}{b} = \frac{a}{b}$.

(B) We know that $x^\circ = \frac{\pi x}{180} \text{ radians}$.
Therefore,$\mathop {\lim }\limits_{x \to 0} \frac{{\sin {x^\circ}}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{\sin \left( \frac{\pi x}{180} \right)}}{x}$.
Multiplying and dividing by $\frac{\pi}{180}$,we get:
$\mathop {\lim }\limits_{x \to 0} \left( \frac{\pi}{180} \right) \cdot \frac{{\sin \left( \frac{\pi x}{180} \right)}}{\frac{\pi x}{180}}$.
Since $\mathop {\lim }\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$,the expression becomes $\frac{\pi}{180} \cdot 1 = \frac{\pi}{180}$.

(D) Using the sum-to-product formulas: $\sin A + \sin B = 2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}$ and $\sin A - \sin B = 2 \cos \frac{A+B}{2} \sin \frac{A-B}{2}$.
$\mathop {\lim }\limits_{x \to 0} \frac{{\sin 2x + \sin 6x}}{{\sin 5x - \sin 3x}} = \mathop {\lim }\limits_{x \to 0} \frac{{2 \sin 4x \cos (-2x)}}{{2 \cos 4x \sin x}} = \mathop {\lim }\limits_{x \to 0} \frac{{\sin 4x \cos 2x}}{{\cos 4x \sin x}}$.
Multiply and divide by $4x$ and $x$ to use the standard limit $\mathop {\lim }\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$:
$= \mathop {\lim }\limits_{x \to 0} \left( \frac{\sin 4x}{4x} \right) \left( \frac{x}{\sin x} \right) \frac{4x \cos 2x}{x \cos 4x} = 1 \times 1 \times 4 \times \frac{\cos(0)}{\cos(0)} = 4 \times 1 = 4$.

(B) We know the standard limit formula $\mathop {\lim }\limits_{x \to 0} \frac{\sin(ax)}{x} = a$.
Applying this to the given expression:
$\mathop {\lim }\limits_{\theta \to 0} \frac{\sin(\theta/4)}{\theta} = \mathop {\lim }\limits_{\theta \to 0} \frac{1}{4} \cdot \frac{\sin(\theta/4)}{\theta/4}$.
Since $\mathop {\lim }\limits_{u \to 0} \frac{\sin(u)}{u} = 1$ where $u = \theta/4$,
The expression becomes $\frac{1}{4} \cdot 1 = \frac{1}{4}$.

(C) We need to evaluate the limit: $\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\tan 3x}}{x} + \cos x} \right)$.
Using the property of limits,$\mathop {\lim }\limits_{x \to a} [f(x) + g(x)] = \mathop {\lim }\limits_{x \to a} f(x) + \mathop {\lim }\limits_{x \to a} g(x)$,we get:
$\mathop {\lim }\limits_{x \to 0} \frac{{\tan 3x}}{x} + \mathop {\lim }\limits_{x \to 0} \cos x$.
For the first term,multiply and divide by $3$:
$\mathop {\lim }\limits_{x \to 0} 3 \cdot \frac{{\tan 3x}}{{3x}} + \mathop {\lim }\limits_{x \to 0} \cos x$.
Since $\mathop {\lim }\limits_{\theta \to 0} \frac{{\tan \theta}}{\theta} = 1$,the first term becomes $3 \cdot 1 = 3$.
For the second term,$\mathop {\lim }\limits_{x \to 0} \cos x = \cos(0) = 1$.
Therefore,the total limit is $3 + 1 = 4$.

(C) Using the trigonometric identity $1 - \cos \theta = 2 \sin^2(\theta/2)$,the limit becomes:
$\mathop {\lim }\limits_{\theta \to 0} \frac{2 \sin^2(\theta/2)}{\theta^2} = \mathop {\lim }\limits_{\theta \to 0} \frac{2 \sin^2(\theta/2)}{4(\theta/2)^2} = \frac{2}{4} \mathop {\lim }\limits_{\theta \to 0} \left( \frac{\sin(\theta/2)}{\theta/2} \right)^2 = \frac{1}{2} \times (1)^2 = \frac{1}{2}$.
Alternatively,applying $L$-Hospital's rule:
$\mathop {\lim }\limits_{\theta \to 0} \frac{1 - \cos \theta}{\theta^2} = \mathop {\lim }\limits_{\theta \to 0} \frac{\sin \theta}{2\theta} = \frac{1}{2} \mathop {\lim }\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = \frac{1}{2} \times 1 = \frac{1}{2}$.

(B) We need to evaluate the limit: $\mathop {\lim }\limits_{\theta \to 0} \frac{{\sin 3\theta - \sin \theta }}{{\sin \theta }}$
Using the formula $\sin 3\theta = 3\sin \theta - 4\sin^3 \theta$:
$= \mathop {\lim }\limits_{\theta \to 0} \frac{{(3\sin \theta - 4\sin^3 \theta) - \sin \theta }}{{\sin \theta }}$
$= \mathop {\lim }\limits_{\theta \to 0} \frac{{2\sin \theta - 4\sin^3 \theta }}{{\sin \theta }}$
$= \mathop {\lim }\limits_{\theta \to 0} (2 - 4\sin^2 \theta)$
$= 2 - 4(0)^2 = 2$
Alternatively,using the property $\mathop {\lim }\limits_{\theta \to 0} \frac{{\sin n\theta }}{{\sin \theta }} = n$:
$= \mathop {\lim }\limits_{\theta \to 0} \frac{{\sin 3\theta }}{{\sin \theta }} - \mathop {\lim }\limits_{\theta \to 0} \frac{{\sin \theta }}{{\sin \theta }}$
$= 3 - 1 = 2$

(A) We evaluate the limit $\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos x}}{x}$.
Using the trigonometric identity $1 - \cos x = 2 \sin^2(x/2)$,the expression becomes:
$\mathop {\lim }\limits_{x \to 0} \frac{2 \sin^2(x/2)}{x}$.
Multiply and divide by $x/2$ to use the standard limit $\mathop {\lim }\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$:
$\mathop {\lim }\limits_{x \to 0} \left( \frac{2 \sin^2(x/2)}{x} \cdot \frac{x/4}{x/4} \right) = \mathop {\lim }\limits_{x \to 0} \left( 2 \cdot \left( \frac{\sin(x/2)}{x/2} \right)^2 \cdot \frac{x}{4} \right) = 2 \cdot (1)^2 \cdot 0 = 0$.
Alternatively,using $L$-Hospital's rule:
$\mathop {\lim }\limits_{x \to 0} \frac{\frac{d}{dx}(1 - \cos x)}{\frac{d}{dx}(x)} = \mathop {\lim }\limits_{x \to 0} \frac{\sin x}{1} = \sin(0) = 0$.

(B) We need to evaluate the limit: $\mathop {\lim }\limits_{x \to 0} \frac{{{x^2} - \tan 2x}}{{\tan x}}$.
Dividing the numerator and denominator by $x$,we get:
$\mathop {\lim }\limits_{x \to 0} \frac{\frac{x^2}{x} - \frac{\tan 2x}{x}}{\frac{\tan x}{x}} = \mathop {\lim }\limits_{x \to 0} \frac{x - 2 \cdot \frac{\tan 2x}{2x}}{\frac{\tan x}{x}}$.
Using the standard limit $\mathop {\lim }\limits_{\theta \to 0} \frac{\tan \theta}{\theta} = 1$,we have:
$\frac{0 - 2(1)}{1} = -2$.

(A) To evaluate the limit $\mathop {\lim }\limits_{\theta \to 0} \frac{5\theta \cos \theta - 2\sin \theta }{3\theta + \tan \theta }$,divide both the numerator and the denominator by $\theta$:
$\mathop {\lim }\limits_{\theta \to 0} \frac{\frac{5\theta \cos \theta - 2\sin \theta}{\theta}}{\frac{3\theta + \tan \theta}{\theta}} = \mathop {\lim }\limits_{\theta \to 0} \frac{5\cos \theta - 2\frac{\sin \theta}{\theta}}{3 + \frac{\tan \theta}{\theta}}$
Using the standard limits $\mathop {\lim }\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$ and $\mathop {\lim }\limits_{\theta \to 0} \frac{\tan \theta}{\theta} = 1$,and knowing $\cos(0) = 1$:
$= \frac{5(1) - 2(1)}{3 + 1} = \frac{5 - 2}{4} = \frac{3}{4}$

(D) We have the limit: $\mathop {\lim }\limits_{h \to 0} \frac{{2\left[ {\sqrt 3 \sin \left( {\frac{\pi }{6} + h} \right) - \cos \left( {\frac{\pi }{6} + h} \right)} \right]}}{{\sqrt 3 h(\sqrt 3 \cos h - \sin h)}}$
Multiply the numerator and denominator by $\frac{1}{2}$ inside the bracket:
$= \mathop {\lim }\limits_{h \to 0} \frac{{4\left[ {\frac{\sqrt 3}{2} \sin \left( {\frac{\pi }{6} + h} \right) - \frac{1}{2} \cos \left( {\frac{\pi }{6} + h} \right)} \right]}}{{\sqrt 3 h(\sqrt 3 \cos h - \sin h)}}$
Using $\sin(A-B) = \sin A \cos B - \cos A \sin B$,where $A = \frac{\pi}{6} + h$ and $B = \frac{\pi}{6}$:
$= \mathop {\lim }\limits_{h \to 0} \frac{{4 \sin \left( {\frac{\pi }{6} + h - \frac{\pi }{6}} \right)}}{{\sqrt 3 h(\sqrt 3 \cos h - \sin h)}}$
$= \mathop {\lim }\limits_{h \to 0} \frac{{4 \sin h}}{{\sqrt 3 h(\sqrt 3 \cos h - \sin h)}}$
$= \frac{4}{\sqrt 3} \times \left( \mathop {\lim }\limits_{h \to 0} \frac{\sin h}{h} \right) \times \left( \mathop {\lim }\limits_{h \to 0} \frac{1}{{\sqrt 3 \cos h - \sin h}} \right)$
$= \frac{4}{\sqrt 3} \times 1 \times \frac{1}{{\sqrt 3(1) - 0}} = \frac{4}{\sqrt 3} \times \frac{1}{\sqrt 3} = \frac{4}{3}$.

(A) We need to evaluate the limit: $\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos x}}{{{{\sin }^2}x}}$.
Using the trigonometric identity $1 - \cos x = 2 \sin^2(\frac{x}{2})$ and $\sin^2 x = (2 \sin(\frac{x}{2}) \cos(\frac{x}{2}))^2 = 4 \sin^2(\frac{x}{2}) \cos^2(\frac{x}{2})$:
$\mathop {\lim }\limits_{x \to 0} \frac{{2 \sin^2(\frac{x}{2})}}{{4 \sin^2(\frac{x}{2}) \cos^2(\frac{x}{2})}} = \mathop {\lim }\limits_{x \to 0} \frac{1}{2 \cos^2(\frac{x}{2})}$.
As $x \to 0$,$\cos(\frac{x}{2}) \to \cos(0) = 1$.
Therefore,the limit is $\frac{1}{2(1)^2} = \frac{1}{2}$.

(C) We are given the limit: $\mathop {\lim }\limits_{x \to 0} \frac{{\sin 3x + \sin x}}{x}$.
Using the property of limits,we can split the expression:
$= \mathop {\lim }\limits_{x \to 0} \frac{\sin 3x}{x} + \mathop {\lim }\limits_{x \to 0} \frac{\sin x}{x}$.
Multiply and divide the first term by $3$:
$= \mathop {\lim }\limits_{x \to 0} 3 \cdot \frac{\sin 3x}{3x} + \mathop {\lim }\limits_{x \to 0} \frac{\sin x}{x}$.
Using the standard limit $\mathop {\lim }\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$:
$= 3(1) + 1 = 3 + 1 = 4$.

(A) We know that $1 - \cos \theta = 2 \sin^2(\theta/2)$.
Therefore,$1 - \cos 6x = 2 \sin^2(3x)$.
$\mathop {\lim }\limits_{x \to 0} \frac{2 \sin^2(3x)}{x} = \mathop {\lim }\limits_{x \to 0} \left( 2 \cdot \frac{\sin^2(3x)}{x} \right)$.
Multiply and divide by $9x$ to use the standard limit $\mathop {\lim }\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$:
$= \mathop {\lim }\limits_{x \to 0} \left( 2 \cdot \frac{\sin^2(3x)}{(3x)^2} \cdot 9x \right) = \mathop {\lim }\limits_{x \to 0} \left( 2 \cdot 1^2 \cdot 9x \right) = 0$.

(B) To evaluate the limit $\mathop {\lim }\limits_{x \to 0} \frac{\sin(mx)}{\tan(nx)}$,we use the standard limits $\mathop {\lim }\limits_{\theta \to 0} \frac{\sin(\theta)}{\theta} = 1$ and $\mathop {\lim }\limits_{\theta \to 0} \frac{\tan(\theta)}{\theta} = 1$.
Divide the numerator and denominator by $x$:
$\mathop {\lim }\limits_{x \to 0} \frac{\frac{\sin(mx)}{x}}{\frac{\tan(nx)}{x}}$
Multiply and divide by $m$ in the numerator and $n$ in the denominator:
$\mathop {\lim }\limits_{x \to 0} \frac{m \cdot \frac{\sin(mx)}{mx}}{n \cdot \frac{\tan(nx)}{nx}}$
As $x \to 0$,$mx \to 0$ and $nx \to 0$,so the limit becomes:
$\frac{m \cdot 1}{n \cdot 1} = \frac{m}{n}$.

(A) We know the trigonometric identity $\sin 3x = 3\sin x - 4\sin^3 x$.
Substituting this into the expression:
$\mathop {\lim }\limits_{x \to 0} \frac{{3\sin x - (3\sin x - 4\sin^3 x)}}{{{x^3}}}$
$= \mathop {\lim }\limits_{x \to 0} \frac{{4\sin^3 x}}{{{x^3}}}$
$= 4 \times \mathop {\lim }\limits_{x \to 0} \left( \frac{\sin x}{x} \right)^3$
Since $\mathop {\lim }\limits_{x \to 0} \frac{\sin x}{x} = 1$,we get:
$= 4 \times (1)^3 = 4$.

(A) We are given the limit: $\mathop {\lim }\limits_{x \to 0} \frac{{{x^3}}}{{\sin {x^2}}}$.
We can rewrite the expression as: $\mathop {\lim }\limits_{x \to 0} \left( \frac{x^2}{\sin {x^2}} \times x \right)$.
Using the property of limits $\mathop {\lim }\limits_{u \to 0} \frac{u}{\sin u} = 1$,where $u = x^2$,as $x \to 0$,$u \to 0$.
So,$\mathop {\lim }\limits_{x \to 0} \frac{x^2}{\sin {x^2}} = 1$.
Therefore,the limit becomes: $1 \times \mathop {\lim }\limits_{x \to 0} x = 1 \times 0 = 0$.

(B) We know that the standard limit formula is $\mathop {\lim }\limits_{x \to 0} \frac{{\tan x}}{x} = 1$.
By taking the reciprocal of the limit,we get:
$\mathop {\lim }\limits_{x \to 0} \frac{x}{{\tan x}} = \frac{1}{{\mathop {\lim }\limits_{x \to 0} \frac{{\tan x}}{x}}} = \frac{1}{1} = 1$.

(A) The limit $\mathop {\lim }\limits_{x \to 0} \frac{\sin x}{x}$ is a standard limit in calculus.
By using the Taylor series expansion for $\sin x$,we have $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots$
Therefore,$\frac{\sin x}{x} = \frac{x - \frac{x^3}{6} + \dots}{x} = 1 - \frac{x^2}{6} + \dots$
As $x \to 0$,the expression approaches $1$.

(C) Using the sum-to-product formula $\sin(A+B) + \sin(A-B) = 2\sin A \cos B$,we have:
$\sin(a+x) + \sin(a-x) = 2\sin a \cos x$.
Substituting this into the limit:
$\mathop {\lim }\limits_{x \to 0} \frac{2\sin a \cos x - 2\sin a}{x \sin x} = \mathop {\lim }\limits_{x \to 0} \frac{2\sin a (\cos x - 1)}{x \sin x}$.
$= -2\sin a \mathop {\lim }\limits_{x \to 0} \left( \frac{1 - \cos x}{x^2} \cdot \frac{x}{\sin x} \right)$.
Since $\mathop {\lim }\limits_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$ and $\mathop {\lim }\limits_{x \to 0} \frac{x}{\sin x} = 1$,we get:
$= -2\sin a \cdot \frac{1}{2} \cdot 1 = -\sin a$.

(A) We are given the limit: $\mathop {\lim }\limits_{a \to 0} \frac{{\sin a - \tan a}}{{{{\sin }^3}a}}$
Rewrite $\tan a$ as $\frac{\sin a}{\cos a}$:
$\mathop {\lim }\limits_{a \to 0} \frac{{\sin a - \frac{\sin a}{\cos a}}}{{{{\sin }^3}a}} = \mathop {\lim }\limits_{a \to 0} \frac{{\sin a(\cos a - 1)}}{{\cos a \cdot \sin^3 a}}$
Simplify by canceling $\sin a$:
$= \mathop {\lim }\limits_{a \to 0} \frac{{\cos a - 1}}{{\cos a \cdot \sin^2 a}}$
Use the identity $\sin^2 a = 1 - \cos^2 a = (1 - \cos a)(1 + \cos a)$:
$= \mathop {\lim }\limits_{a \to 0} \frac{{-(\cos a - 1)}}{{\cos a \cdot (1 - \cos a)(1 + \cos a)}}$
Cancel $(1 - \cos a)$:
$= \mathop {\lim }\limits_{a \to 0} \frac{{-1}}{{\cos a(1 + \cos a)}}$
Substitute $a = 0$:
$= \frac{{-1}}{{1(1 + 1)}} = -\frac{1}{2}$

(C) We know that $\mathop {\lim }\limits_{u \to 0} \frac{k^u - 1}{u} = \ln k$.
Given the limit: $\mathop {\lim }\limits_{x \to 0} \frac{a^{\sin x} - 1}{b^{\sin x} - 1}$.
Multiply and divide by $\sin x$: $\mathop {\lim }\limits_{x \to 0} \left( \frac{a^{\sin x} - 1}{\sin x} \times \frac{\sin x}{b^{\sin x} - 1} \right)$.
As $x \to 0$,$\sin x \to 0$. Thus,the expression becomes $\ln a \times \frac{1}{\ln b}$.
Therefore,the result is $\frac{\log a}{\log b}$.

(A) Given limit: $\mathop {\lim }\limits_{x \to 0} \frac{(1 - \cos 2x)\sin 5x}{x^2 \sin 3x}$
Using the identity $1 - \cos 2x = 2 \sin^2 x$,we get:
$= \mathop {\lim }\limits_{x \to 0} \frac{2 \sin^2 x \sin 5x}{x^2 \sin 3x}$
$= \mathop {\lim }\limits_{x \to 0} 2 \left( \frac{\sin x}{x} \right)^2 \cdot \frac{\sin 5x}{\sin 3x}$
Multiply and divide by $5x$ and $3x$:
$= \mathop {\lim }\limits_{x \to 0} 2 \left( \frac{\sin x}{x} \right)^2 \cdot \frac{\frac{\sin 5x}{5x} \cdot 5x}{\frac{\sin 3x}{3x} \cdot 3x}$
Since $\mathop {\lim }\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$:
$= 2 \cdot (1)^2 \cdot \frac{1 \cdot 5}{1 \cdot 3} = \frac{10}{3}$.

(B) We need to evaluate the limit $L = \mathop {\lim }\limits_{x \to 0} \frac{{\sin (\pi {{\cos }^2}x)}}{{{x^2}}}$.
Since $\cos^2 x = 1 - \sin^2 x$,we can rewrite the expression as:
$L = \mathop {\lim }\limits_{x \to 0} \frac{{\sin (\pi (1 - \sin^2 x))}}{{{x^2}}}$
$L = \mathop {\lim }\limits_{x \to 0} \frac{{\sin (\pi - \pi \sin^2 x)}}{{{x^2}}}$
Using the identity $\sin(\pi - \theta) = \sin \theta$,we get:
$L = \mathop {\lim }\limits_{x \to 0} \frac{{\sin (\pi \sin^2 x)}}{{{x^2}}}$
Multiply and divide by $\pi \sin^2 x$:
$L = \mathop {\lim }\limits_{x \to 0} \left( \frac{{\sin (\pi \sin^2 x)}}{{\pi \sin^2 x}} \right) \times \frac{{\pi \sin^2 x}}{{{x^2}}}$
As $x \to 0$,$\pi \sin^2 x \to 0$,so $\frac{{\sin (\pi \sin^2 x)}}{{\pi \sin^2 x}} \to 1$.
Also,$\mathop {\lim }\limits_{x \to 0} \frac{{\sin^2 x}}{{{x^2}}} = 1$.
Therefore,$L = 1 \times \pi \times 1 = \pi $.

(D) We need to evaluate the limit: $\mathop {\lim }\limits_{x \to 0} \frac{{\cos (\sin x) - 1}}{{{x^2}}}$
Using the trigonometric identity $\cos \theta - 1 = -2 \sin^2(\theta/2)$,we substitute $\theta = \sin x$:
$= \mathop {\lim }\limits_{x \to 0} \frac{{ - 2 \sin^2(\frac{\sin x}{2})}}{{{x^2}}}$
$= -2 \mathop {\lim }\limits_{x \to 0} \left( \frac{\sin(\frac{\sin x}{2})}{\frac{\sin x}{2}} \cdot \frac{\sin x}{2x} \right)^2$
Since $\mathop {\lim }\limits_{u \to 0} \frac{\sin u}{u} = 1$ and $\mathop {\lim }\limits_{x \to 0} \frac{\sin x}{x} = 1$,we get:
$= -2 \cdot (1)^2 \cdot (1/2)^2 = -2 \cdot \frac{1}{4} = -\frac{1}{2}$.

(B) We have $\mathop {\lim }\limits_{\theta \to 0} \frac{{4\theta (\tan \theta - \sin \theta )}}{{{{(1 - \cos 2\theta )}^2}}}$.
Using $\tan \theta - \sin \theta = \sin \theta (\sec \theta - 1) = \frac{\sin \theta (1 - \cos \theta )}{\cos \theta }$ and $1 - \cos 2\theta = 2\sin^2 \theta$,the expression becomes:
$\mathop {\lim }\limits_{\theta \to 0} \frac{{4\theta \sin \theta (1 - \cos \theta )}}{{\cos \theta (2\sin^2 \theta )^2}} = \mathop {\lim }\limits_{\theta \to 0} \frac{{4\theta \sin \theta (2\sin^2(\theta /2))}}{{4\sin^4 \theta \cos \theta }}$
$= \mathop {\lim }\limits_{\theta \to 0} \frac{{2\theta \sin^2(\theta /2)}}{{\sin^4 \theta \cos \theta }} = \mathop {\lim }\limits_{\theta \to 0} \frac{{2\theta \sin^2(\theta /2)}}{{(2\sin(\theta /2)\cos(\theta /2))^4 \cos \theta }}$
$= \mathop {\lim }\limits_{\theta \to 0} \frac{{2\theta \sin^2(\theta /2)}}{{16\sin^4(\theta /2)\cos^4(\theta /2) \cos \theta }} = \mathop {\lim }\limits_{\theta \to 0} \frac{\theta }{{8\sin^2(\theta /2)\cos^4(\theta /2) \cos \theta }}$
Using $\sin(\theta /2) \approx \theta /2$,we get $\frac{\theta }{{8(\theta^2/4)(1)(1)}} = \frac{\theta }{{2\theta^2}} = \frac{1}{{2\theta}}$. Wait,let us re-evaluate.
$\mathop {\lim }\limits_{\theta \to 0} \frac{{4\theta \sin \theta (1 - \cos \theta )}}{{4\sin^4 \theta \cos \theta }} = \mathop {\lim }\limits_{\theta \to 0} \frac{{\theta \sin \theta (2\sin^2(\theta /2))}}{{\sin^4 \theta \cos \theta }} = \mathop {\lim }\limits_{\theta \to 0} \frac{{2\theta \sin^2(\theta /2)}}{{\sin^3 \theta \cos \theta }}$
$= \mathop {\lim }\limits_{\theta \to 0} \frac{{2\theta (\theta /2)^2}}{{\theta^3 \cos \theta }} = \mathop {\lim }\limits_{\theta \to 0} \frac{{2\theta^3 / 4}}{{\theta^3}} = \frac{1}{2}$.

(C) Let $L = \mathop {\lim }\limits_{x \to 1} (1 - x)\tan \left( {\frac{{\pi x}}{2}} \right)$.
Substitute $y = 1 - x$,so as $x \to 1$,$y \to 0$ and $x = 1 - y$.
Then $L = \mathop {\lim }\limits_{y \to 0} y \tan \left( {\frac{{\pi (1 - y)}}{2}} \right) = \mathop {\lim }\limits_{y \to 0} y \tan \left( {\frac{\pi }{2} - \frac{{\pi y}}{2}} \right)$.
Using the identity $\tan \left( {\frac{\pi }{2} - \theta } \right) = \cot \theta$,we get:
$L = \mathop {\lim }\limits_{y \to 0} y \cot \left( {\frac{{\pi y}}{2}} \right) = \mathop {\lim }\limits_{y \to 0} \frac{y}{\tan \left( {\frac{{\pi y}}{2}} \right)}$.
Multiply and divide by $\frac{\pi }{2}$:
$L = \mathop {\lim }\limits_{y \to 0} \frac{1}{\frac{\pi }{2}} \cdot \frac{{\frac{{\pi y}}{2}}}{{\tan \left( {\frac{{\pi y}}{2}} \right)}} = \frac{2}{\pi } \cdot 1 = \frac{2}{\pi }$.

(D) Let $f(x) = \frac{\sqrt{1 - \cos\{2(x - 2)\}}}{x - 2}$.
Using the identity $1 - \cos(2\theta) = 2\sin^2(\theta)$,we get $1 - \cos\{2(x - 2)\} = 2\sin^2(x - 2)$.
Thus,$f(x) = \frac{\sqrt{2\sin^2(x - 2)}}{x - 2} = \frac{\sqrt{2}|\sin(x - 2)|}{x - 2}$.
For the Left Hand Limit $(LHL)$: $\mathop {\lim }\limits_{x \to 2^-} \frac{\sqrt{2}|\sin(x - 2)|}{x - 2}$. As $x \to 2^-$,$(x - 2) < 0$,so $|\sin(x - 2)| = -\sin(x - 2)$.
$LHL = \mathop {\lim }\limits_{x \to 2^-} \frac{-\sqrt{2}\sin(x - 2)}{x - 2} = -\sqrt{2}(1) = -\sqrt{2}$.
For the Right Hand Limit $(RHL)$: $\mathop {\lim }\limits_{x \to 2^+} \frac{\sqrt{2}|\sin(x - 2)|}{x - 2}$. As $x \to 2^+$,$(x - 2) > 0$,so $|\sin(x - 2)| = \sin(x - 2)$.
$RHL = \mathop {\lim }\limits_{x \to 2^+} \frac{\sqrt{2}\sin(x - 2)}{x - 2} = \sqrt{2}(1) = \sqrt{2}$.
Since $LHL \neq RHL$,the limit does not exist.

(C) Let $L = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\cot x - \cos x}}{{{{\left( {\pi - 2x} \right)}^3}}}$.
We can rewrite the expression as:
$L = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\cos x \left( {\frac{1}{{\sin x}} - 1} \right)}}{{ - 8{{\left( {x - \frac{\pi }{2}} \right)}^3}}} = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\cos x(1 - \sin x)}}{{\sin x \cdot 8{{\left( {\frac{\pi }{2} - x} \right)}^3}}}$.
Substitute $t = \frac{\pi }{2} - x$. As $x \to \frac{\pi }{2}$,$t \to 0$. Then $x = \frac{\pi }{2} - t$ and $\cos x = \sin t$,$\sin x = \cos t$.
$L = \mathop {\lim }\limits_{t \to 0} \frac{{\sin t(1 - \cos t)}}{{8{t^3}\cos t}}$.
$L = \frac{1}{8} \cdot \mathop {\lim }\limits_{t \to 0} \left( \frac{{\sin t}}{t} \right) \cdot \left( \frac{{1 - \cos t}}{{{t^2}}} \right) \cdot \frac{1}{{\cos t}}$.
Using standard limits $\mathop {\lim }\limits_{t \to 0} \frac{{\sin t}}{t} = 1$ and $\mathop {\lim }\limits_{t \to 0} \frac{{1 - \cos t}}{{{t^2}}} = \frac{1}{2}$:
$L = \frac{1}{8} \cdot 1 \cdot \frac{1}{2} \cdot 1 = \frac{1}{{16}}$.

(B) Let $L = \lim_{x \to 0} (\cos ax)^{\csc^2 bx}$. This is of the form $1^\infty$.
Using the formula $\lim_{x \to 0} f(x)^{g(x)} = e^{\lim_{x \to 0} g(x)(f(x)-1)}$,we have:
$L = e^{\lim_{x \to 0} \csc^2 bx (\cos ax - 1)}$
$L = e^{-\lim_{x \to 0} \frac{1 - \cos ax}{\sin^2 bx}}$
Using the identity $1 - \cos \theta = 2\sin^2(\theta/2)$ and $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$:
$L = e^{-\lim_{x \to 0} \frac{2\sin^2(ax/2)}{\sin^2 bx}} = e^{-\lim_{x \to 0} \frac{2(ax/2)^2}{(bx)^2}} = e^{-\frac{2(a^2/4)}{b^2}} = e^{-\frac{a^2}{2b^2}}$

(D) For a small positive angle $\theta$ (in radians),we consider the unit circle geometry.
In the first quadrant,for $0 < \theta < \frac{\pi}{2}$,the following inequality holds: $\sin \theta < \theta < \tan \theta$.
Dividing the entire inequality by $\sin \theta$ (which is positive),we get $1 < \frac{\theta}{\sin \theta} < \frac{1}{\cos \theta}$.
Taking the reciprocal,we get $\cos \theta < \frac{\sin \theta}{\theta} < 1$.
Also,dividing the original inequality $\sin \theta < \theta < \tan \theta$ by $\theta$,we get $\frac{\sin \theta}{\theta} < 1 < \frac{\tan \theta}{\theta}$.
This implies $\frac{\tan \theta}{\theta} > 1 > \frac{\sin \theta}{\theta}$,which confirms that $\frac{\tan \theta}{\theta} > \frac{\sin \theta}{\theta}$.
Thus,both statements $(B)$ and $(C)$ are correct.

(C) Let the expression be $L = \mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - 1}}{{{{\sin }^2}x + \cos x \cos (x + 2) - {{\cos }^2}(x + 1)}}$.
Using the identity $\cos A \cos B = \frac{1}{2} [\cos(A+B) + \cos(A-B)]$,we have $\cos x \cos (x+2) = \frac{1}{2} [\cos(2x+2) + \cos(-2)] = \frac{1}{2} [\cos(2x+2) + \cos 2]$.
Also,${{\sin }^2}x - {{\cos }^2}(x+1) = \frac{1 - \cos 2x}{2} - \frac{1 + \cos(2x+2)}{2} = \frac{-\cos 2x - \cos(2x+2)}{2}$.
Substituting these,the denominator becomes $\frac{1}{2} [-\cos 2x - \cos(2x+2) + \cos(2x+2) + \cos 2] = \frac{\cos 2 - \cos 2x}{2} = \frac{2 \sin(\frac{2x+2}{2}) \sin(\frac{2x-2}{2})}{2} = \sin(x+1) \sin(x-1)$.
Thus,$L = \mathop {\lim }\limits_{x \to 1} \frac{(x-1)(x+1)}{\sin(x+1) \sin(x-1)} = \mathop {\lim }\limits_{x \to 1} \left( \frac{x-1}{\sin(x-1)} \right) \cdot \frac{x+1}{\sin(x+1)} = 1 \cdot \frac{2}{\sin 2} = \frac{2}{\sin 2}$.

(A) The given limit is of the form $1^\infty$.
Using the formula $\mathop {\lim }\limits_{x \to 0} {f(x)^{g(x)}} = e^{\mathop {\lim }\limits_{x \to 0} (f(x) - 1)g(x)}$,we have:
$L = e^{\mathop {\lim }\limits_{x \to 0} (\cos ax - 1) \csc^2 bx}$
$L = e^{\mathop {\lim }\limits_{x \to 0} \frac{-(1 - \cos ax)}{\sin^2 bx}}$
$L = e^{\mathop {\lim }\limits_{x \to 0} -\left( \frac{1 - \cos ax}{(ax)^2} \right) \cdot \frac{(ax)^2}{\left( \frac{\sin bx}{bx} \cdot bx \right)^2}}$
Since $\mathop {\lim }\limits_{\theta \to 0} \frac{1 - \cos \theta}{\theta^2} = \frac{1}{2}$ and $\mathop {\lim }\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$:
$L = e^{-\frac{1}{2} \cdot \frac{a^2}{b^2}} = e^{\left( -\frac{a^2}{2b^2} \right)}$

(A) Given the limit: $L = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\sqrt {1 - \cos 2ax} }}{{\sin bx}}$
Using the identity $1 - \cos 2\theta = 2\sin^2 \theta$,we have:
$L = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\sqrt {2\sin^2 ax} }}{{\sin bx}} = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{\sqrt{2} |\sin ax|}}{{\sin bx}}$
Since $x \to 0^+$,$ax > 0$ (as $a > 0$),so $|\sin ax| = \sin ax$.
$L = \sqrt{2} \mathop {\lim }\limits_{x \to {0^ + }} \frac{\sin ax}{\sin bx} = \sqrt{2} \mathop {\lim }\limits_{x \to {0^ + }} \left( \frac{\sin ax}{ax} \cdot \frac{bx}{\sin bx} \cdot \frac{ax}{bx} \right)$
$L = \sqrt{2} \cdot 1 \cdot 1 \cdot \frac{a}{b} = \frac{a\sqrt{2}}{b}$

(B) To evaluate the limit $\mathop {\lim }\limits_{x \to 0} \frac{{5\sin x + x\cos x}}{{2\tan x - {x^2}}}$,we divide both the numerator and the denominator by $x$:
$\mathop {\lim }\limits_{x \to 0} \frac{{\frac{5\sin x}{x} + \frac{x\cos x}{x}}}{{\frac{2\tan x}{x} - \frac{x^2}{x}}}$
$= \mathop {\lim }\limits_{x \to 0} \frac{{5(\frac{\sin x}{x}) + \cos x}}{{2(\frac{\tan x}{x}) - x}}$
Using the standard limits $\mathop {\lim }\limits_{x \to 0} \frac{\sin x}{x} = 1$ and $\mathop {\lim }\limits_{x \to 0} \frac{\tan x}{x} = 1$,we get:
$= \frac{5(1) + \cos(0)}{2(1) - 0} = \frac{5 + 1}{2} = \frac{6}{2} = 3$

(D) Let $L = \mathop {\lim }\limits_{x \to 0} \frac{{x\tan 2x - 2x\tan x}}{{{{\left( {1 - \cos 2x} \right)}^2}}}$.
Using the identity $\tan 2x = \frac{{2\tan x}}{{1 - {{\tan }^2}x}}$ and $1 - \cos 2x = 2\sin^2 x$,we have:
$L = \mathop {\lim }\limits_{x \to 0} \frac{{x\left( \frac{{2\tan x}}{{1 - {{\tan }^2}x}} \right) - 2x\tan x}}{{{{\left( {2\sin^2 x} \right)}^2}}}$
$L = \mathop {\lim }\limits_{x \to 0} \frac{{2x\tan x - 2x\tan x(1 - {{\tan }^2}x)}}{{(1 - {{\tan }^2}x) \cdot 4\sin^4 x}}$
$L = \mathop {\lim }\limits_{x \to 0} \frac{{2x\tan^3 x}}{{4\sin^4 x(1 - {{\tan }^2}x)}}$
Since $\tan x = \frac{\sin x}{\cos x}$,we get:
$L = \mathop {\lim }\limits_{x \to 0} \frac{{2x \cdot \frac{\sin^3 x}{\cos^3 x}}}{{4\sin^4 x(1 - {{\tan }^2}x)}}$
$L = \mathop {\lim }\limits_{x \to 0} \frac{{2x}}{{4\sin x \cdot \cos^3 x(1 - {{\tan }^2}x)}}$
$L = \mathop {\lim }\limits_{x \to 0} \left( \frac{x}{2\sin x} \right) \cdot \mathop {\lim }\limits_{x \to 0} \frac{1}{{\cos^3 x(1 - {{\tan }^2}x)}}$
$L = \frac{1}{2} \cdot \frac{1}{1(1 - 0)} = \frac{1}{2}$.

(D) We need to evaluate the limit: $\mathop {\lim }\limits_{x \to 0} \frac{{\sin \left( {\pi {{\cos }^2}x} \right)}}{{{x^2}}}$
Using the identity $\cos^2 x = 1 - \sin^2 x$,we rewrite the expression:
$= \mathop {\lim }\limits_{x \to 0} \frac{{\sin \left( {\pi (1 - \sin^2 x)} \right)}}{{{x^2}}}$
$= \mathop {\lim }\limits_{x \to 0} \frac{{\sin \left( {\pi - \pi \sin^2 x} \right)}}{{{x^2}}}$
Since $\sin(\pi - \theta) = \sin \theta$,we have:
$= \mathop {\lim }\limits_{x \to 0} \frac{{\sin \left( {\pi \sin^2 x} \right)}}{{{x^2}}}$
Multiply and divide by $\pi \sin^2 x$:
$= \mathop {\lim }\limits_{x \to 0} \left( \frac{{\sin \left( {\pi \sin^2 x} \right)}}{{\pi \sin^2 x}} \times \frac{{\pi \sin^2 x}}{{{x^2}}} \right)$
Using the standard limit $\mathop {\lim }\limits_{\theta \to 0} \frac{{\sin \theta}}{\theta} = 1$ and $\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1$:
$= 1 \times \pi \times (\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x})^2 = 1 \times \pi \times 1^2 = \pi$

(D) Given limit: $L = \mathop {\lim }\limits_{x \to 0} \frac{x \cot(4x)}{\sin^2 x \cot^2(2x)}$
Substitute $\cot \theta = \frac{\cos \theta}{\sin \theta}$:
$L = \mathop {\lim }\limits_{x \to 0} \frac{x \cos(4x) \sin^2(2x)}{\sin^2 x \sin(4x) \cos^2(2x)}$
Use the identity $\sin(4x) = 2 \sin(2x) \cos(2x)$:
$L = \mathop {\lim }\limits_{x \to 0} \frac{x \cos(4x) \sin^2(2x)}{\sin^2 x (2 \sin(2x) \cos(2x)) \cos^2(2x)}$
Simplify the expression:
$L = \mathop {\lim }\limits_{x \to 0} \left( \frac{x}{\sin x} \right)^2 \cdot \frac{\sin(2x)}{2x} \cdot \frac{\cos(4x)}{\cos^3(2x)}$
As $x \to 0$,$\frac{\sin \theta}{\theta} \to 1$ and $\cos \theta \to 1$:
$L = (1)^2 \cdot 1 \cdot \frac{1}{1^3} = 1$

(B) We know that $\mathop {\lim }\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$.
$\mathop {\lim }\limits_{x \to 0} \frac{\sin 4x}{\sin 2x} = \mathop {\lim }\limits_{x \to 0} \left( \frac{\sin 4x}{4x} \cdot 4x \cdot \frac{1}{\frac{\sin 2x}{2x} \cdot 2x} \right)$
$= \mathop {\lim }\limits_{x \to 0} \left( \frac{\sin 4x}{4x} \cdot \frac{2x}{\sin 2x} \cdot \frac{4x}{2x} \right)$
$= \left( \mathop {\lim }\limits_{4x \to 0} \frac{\sin 4x}{4x} \right) \cdot \left( \mathop {\lim }\limits_{2x \to 0} \frac{2x}{\sin 2x} \right) \cdot 2$
$= 1 \cdot 1 \cdot 2 = 2$.

(A) We have $\mathop {\lim }\limits_{x \to 0} \frac{\tan x}{x} = \mathop {\lim }\limits_{x \to 0} \frac{\sin x}{x \cos x}$
$= \left( \mathop {\lim }\limits_{x \to 0} \frac{\sin x}{x} \right) \times \left( \mathop {\lim }\limits_{x \to 0} \frac{1}{\cos x} \right)$
$= 1 \times \frac{1}{\cos(0)} = 1 \times 1 = 1$

(A) We are given the limit: $\mathop {\lim }\limits_{x \to 0} \frac{\sin ax}{bx}$
At $x=0$,the function takes the indeterminate form $\frac{0}{0}$.
To evaluate this,we use the standard limit formula $\mathop {\lim }\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$.
Multiply and divide by $a$:
$\mathop {\lim }\limits_{x \to 0} \frac{\sin ax}{bx} = \mathop {\lim }\limits_{x \to 0} \left( \frac{\sin ax}{ax} \times \frac{ax}{bx} \right)$
$= \mathop {\lim }\limits_{x \to 0} \left( \frac{\sin ax}{ax} \right) \times \frac{a}{b}$
Since $x \to 0$,it follows that $ax \to 0$. Therefore:
$= \frac{a}{b} \times \mathop {\lim }\limits_{ax \to 0} \left( \frac{\sin ax}{ax} \right)$
$= \frac{a}{b} \times 1 = \frac{a}{b}$

(A) We are given the limit: $\mathop {\lim }\limits_{x \to 0} \frac{\sin ax}{\sin bx}$.
At $x=0$,the expression takes the indeterminate form $\frac{0}{0}$.
We use the standard limit $\mathop {\lim }\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$.
$\mathop {\lim }\limits_{x \to 0} \frac{\sin ax}{\sin bx} = \mathop {\lim }\limits_{x \to 0} \left( \frac{\sin ax}{ax} \cdot ax \cdot \frac{1}{\frac{\sin bx}{bx} \cdot bx} \right)$
$= \mathop {\lim }\limits_{x \to 0} \left( \frac{\sin ax}{ax} \right) \cdot \frac{1}{\mathop {\lim }\limits_{x \to 0} \left( \frac{\sin bx}{bx} \right)} \cdot \frac{ax}{bx}$
$= 1 \cdot \frac{1}{1} \cdot \frac{a}{b} = \frac{a}{b}$.

(A) We are given the limit: $\mathop {\lim }\limits_{x \to \pi } \frac{\sin (\pi-x)}{\pi(\pi-x)}$
Let $y = \pi - x$. As $x \to \pi$,$y \to 0$.
Substituting this into the limit,we get:
$\mathop {\lim }\limits_{y \to 0} \frac{\sin y}{\pi y}$
$= \frac{1}{\pi} \mathop {\lim }\limits_{y \to 0} \frac{\sin y}{y}$
Using the standard limit formula $\mathop {\lim }\limits_{y \to 0} \frac{\sin y}{y} = 1$,we have:
$= \frac{1}{\pi} \times 1 = \frac{1}{\pi}$