Limit of trigonometric function Questions in English

(B) We need to evaluate $\mathop {\lim }\limits_{x \to 0} \frac{\cos 2x - 1}{\cos x - 1}$.
At $x = 0$,the expression takes the indeterminate form $\frac{0}{0}$.
Using the trigonometric identity $\cos 2\theta = 1 - 2\sin^2 \theta$,we have:
$\cos 2x = 1 - 2\sin^2 x$ and $\cos x = 1 - 2\sin^2 \frac{x}{2}$.
Substituting these into the limit:
$\mathop {\lim }\limits_{x \to 0} \frac{(1 - 2\sin^2 x) - 1}{(1 - 2\sin^2 \frac{x}{2}) - 1} = \mathop {\lim }\limits_{x \to 0} \frac{-2\sin^2 x}{-2\sin^2 \frac{x}{2}} = \mathop {\lim }\limits_{x \to 0} \frac{\sin^2 x}{\sin^2 \frac{x}{2}}$.
Dividing numerator and denominator by $x^2$ and multiplying by appropriate factors:
$= \mathop {\lim }\limits_{x \to 0} \frac{(\frac{\sin x}{x})^2}{(\frac{\sin(x/2)}{x/2})^2 \cdot \frac{1}{4}} = \frac{1^2}{1^2 \cdot \frac{1}{4}} = 4$.

(A) Given the limit: $\mathop {\lim }\limits_{x \to 0} \frac{\sin ax + bx}{ax + \sin bx}$.
At $x=0$,the expression takes the indeterminate form $\frac{0}{0}$.
We can divide the numerator and the denominator by $x$:
$\mathop {\lim }\limits_{x \to 0} \frac{\frac{\sin ax}{x} + b}{\frac{ax}{x} + \frac{\sin bx}{x}}$
Using the standard limit $\mathop {\lim }\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$,we multiply and divide by $a$ and $b$ respectively:
$= \mathop {\lim }\limits_{x \to 0} \frac{a \cdot \frac{\sin ax}{ax} + b}{a + b \cdot \frac{\sin bx}{bx}}$
Applying the limit as $x \to 0$:
$= \frac{a(1) + b}{a + b(1)}$
$= \frac{a+b}{a+b}$
$= 1$.

(A) We are given the limit $L = \lim _{\theta \rightarrow 0} \frac{\tan (\pi \cos ^{2} \theta)}{\sin (2 \pi \sin ^{2} \theta)}$.
Using the identity $\cos^2 \theta = 1 - \sin^2 \theta$,we get $\tan (\pi \cos^2 \theta) = \tan (\pi - \pi \sin^2 \theta)$.
Since $\tan (\pi - x) = -\tan x$,we have $\tan (\pi - \pi \sin^2 \theta) = -\tan (\pi \sin^2 \theta)$.
Substituting this into the limit,we get $L = \lim _{\theta \rightarrow 0} \frac{-\tan (\pi \sin^2 \theta)}{\sin (2 \pi \sin^2 \theta)}$.
As $\theta \rightarrow 0$,$\sin^2 \theta \rightarrow 0$. Let $u = \pi \sin^2 \theta$,then as $\theta \rightarrow 0$,$u \rightarrow 0$.
The expression becomes $\lim _{u \rightarrow 0} \frac{-\tan u}{\sin (2u)}$.
Using the standard limits $\lim _{x \rightarrow 0} \frac{\tan x}{x} = 1$ and $\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1$,we multiply and divide by $u$ and $2u$:
$L = \lim _{u}$ ${\rightarrow 0} -\left( \frac{\tan u}{u} \right) \left( \frac{2u}{\sin (2u)} \right) \times \frac{1}{2} = -1 \times 1 \times \frac{1}{2} = -\frac{1}{2}$.

(C) We need to evaluate $L = \lim _{x \rightarrow 0} \frac{\sin ^{2}\left(\pi \cos ^{4} x\right)}{x^{4}}$.
Since $\cos x \rightarrow 1$ as $x \rightarrow 0$,the argument $\pi \cos^4 x \rightarrow \pi$.
Let $u = \pi \cos^4 x$. Then $\sin^2(u) = \sin^2(\pi - u) = \sin^2(\pi(1 - \cos^4 x))$.
As $x \rightarrow 0$,$1 - \cos^4 x = (1 - \cos^2 x)(1 + \cos^2 x) = \sin^2 x (1 + \cos^2 x)$.
So,$L = \lim _{x \rightarrow 0} \frac{\sin^2(\pi \sin^2 x (1 + \cos^2 x))}{x^4}$.
Using the limit $\lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1$,we have:
$L = \lim _{x}$ ${\rightarrow 0} \left[ \frac{\sin(\pi \sin^2 x (1 + \cos^2 x))}{\pi \sin^2 x (1 + \cos^2 x)} \right]^2 \cdot \frac{\pi^2 \sin^4 x (1 + \cos^2 x)^2}{x^4}$.
$L = 1^2 \cdot \pi^2 \cdot \lim_{x \rightarrow 0} \left( \frac{\sin x}{x} \right)^4 \cdot (1 + \cos^2 x)^2$.
$L = \pi^2 \cdot (1)^4 \cdot (1 + 1)^2 = \pi^2 \cdot 4 = 4 \pi^2$.

(A) Let $p = \lim_{x \rightarrow 0} \left(\frac{x}{\sin x}\right)^{6/x^2}$.
Taking the natural logarithm on both sides:
$\ln p = \lim_{x \rightarrow 0} \frac{6}{x^2} \ln \left(\frac{x}{\sin x}\right)$.
Using the Taylor series expansion for $\sin x = x - \frac{x^3}{6} + O(x^5)$:
$\frac{x}{\sin x} = \frac{x}{x - \frac{x^3}{6} + O(x^5)} = \left(1 - \frac{x^2}{6} + O(x^4)\right)^{-1} = 1 + \frac{x^2}{6} + O(x^4)$.
Now,$\ln p = \lim_{x \rightarrow 0} \frac{6}{x^2} \ln \left(1 + \frac{x^2}{6} + O(x^4)\right)$.
Using the expansion $\ln(1+u) = u - \frac{u^2}{2} + \dots$ for $u = \frac{x^2}{6}$:
$\ln p = \lim_{x \rightarrow 0} \frac{6}{x^2} \left(\frac{x^2}{6} + O(x^4)\right) = \lim_{x \rightarrow 0} (1 + O(x^2)) = 1$.
Since $\ln p = 1$,we have $p = e^1 = e$.

(B) Let $L = \lim _{t}$ ${\rightarrow 0}\left(1^{\operatorname{cosec}^2 t}+2^{\operatorname{cosec}^2 t}+\ldots +n^{\operatorname{cosec}^2 t}\right)^{\sin ^2 t}$.
We can factor out the largest term $n^{\operatorname{cosec}^2 t}$ from the bracket:
$L = \lim _{t}$ ${\rightarrow 0} \left[ n^{\operatorname{cosec}^2 t} \left( \left(\frac{1}{n}\right)^{\operatorname{cosec}^2 t} + \left(\frac{2}{n}\right)^{\operatorname{cosec}^2 t} + \ldots + 1 \right) \right]^{\sin ^2 t}$.
$L = \lim _{t}$ ${\rightarrow 0} \left( n^{\operatorname{cosec}^2 t} \right)^{\sin ^2 t} \cdot \left( \left(\frac{1}{n}\right)^{\operatorname{cosec}^2 t} + \left(\frac{2}{n}\right)^{\operatorname{cosec}^2 t} + \ldots + 1 \right)^{\sin ^2 t}$.
Since $\operatorname{cosec}^2 t \cdot \sin ^2 t = 1$,the first part becomes $n^1 = n$.
For the second part,as $t \rightarrow 0$,$\operatorname{cosec}^2 t \rightarrow \infty$. Thus,for $k < n$,$\left(\frac{k}{n}\right)^{\operatorname{cosec}^2 t} \rightarrow 0$.
So,the expression becomes $n \cdot (0 + 0 + \ldots + 1)^0 = n \cdot 1^0 = n \cdot 1 = n$.

(B) Let $L = \lim _{x \rightarrow 0} \left( \frac{1-\cos ^2(3 x)}{\cos ^3(4 x)} \cdot \frac{\sin ^3(4 x)}{(\ln(1+2 x))^5} \right)$.
Using the identity $1-\cos^2 \theta = \sin^2 \theta$,we have:
$L = \lim _{x \rightarrow 0} \frac{\sin^2(3x)}{\cos^3(4x)} \cdot \frac{\sin^3(4x)}{(\ln(1+2x))^5}$.
Multiply and divide by the necessary terms to use standard limits $\lim_{u \to 0} \frac{\sin u}{u} = 1$ and $\lim_{u \to 0} \frac{\ln(1+u)}{u} = 1$:
$L = \lim _{x}$ ${\rightarrow 0} \left( \frac{\sin^2(3x)}{(3x)^2} \cdot (3x)^2 \right) \cdot \frac{1}{\cos^3(4x)} \cdot \left( \frac{\sin^3(4x)}{(4x)^3} \cdot (4x)^3 \right) \cdot \left( \frac{2x}{\ln(1+2x)} \right)^5 \cdot \frac{1}{(2x)^5}$.
$L = \lim _{x}$ ${\rightarrow 0} \left( 1^2 \cdot 9x^2 \right) \cdot \frac{1}{1} \cdot \left( 1^3 \cdot 64x^3 \right) \cdot 1^5 \cdot \frac{1}{32x^5}$.
$L = \lim _{x \rightarrow 0} \frac{9 \cdot 64 \cdot x^5}{32 \cdot x^5} = \frac{576}{32} = 18$.

(C) Let the given limit be $L = \lim_{x \rightarrow \frac{\pi}{2}} \frac{4 \sqrt{2}(\sin 3x + \sin x)}{\left(2 \sin 2x \sin \frac{3x}{2} + \cos \frac{5x}{2}\right) - \left(\sqrt{2} + \sqrt{2} \cos 2x + \cos \frac{3x}{2}\right)}$.
Using the identity $\sin 3x + \sin x = 2 \sin 2x \cos x$,the numerator becomes $8 \sqrt{2} \sin 2x \cos x = 16 \sqrt{2} \sin x \cos^2 x$.
For the denominator,we use $\cos \frac{5x}{2} - \cos \frac{3x}{2} = -2 \sin 2x \sin \frac{x}{2}$ and $\sqrt{2} + \sqrt{2} \cos 2x = \sqrt{2}(1 + \cos 2x) = 2 \sqrt{2} \cos^2 x$.
The denominator simplifies to $2 \sin 2x \sin \frac{3x}{2} - 2 \sin 2x \sin \frac{x}{2} - 2 \sqrt{2} \cos^2 x = 2 \sin 2x (\sin \frac{3x}{2} - \sin \frac{x}{2}) - 2 \sqrt{2} \cos^2 x$.
Using $\sin \frac{3x}{2} - \sin \frac{x}{2} = 2 \cos x \sin \frac{x}{2}$,the denominator becomes $4 \sin x \cos x (2 \cos x \sin \frac{x}{2}) - 2 \sqrt{2} \cos^2 x = 2 \cos^2 x (4 \sin x \sin \frac{x}{2} - \sqrt{2})$.
Thus,$L = \lim_{x \rightarrow \frac{\pi}{2}} \frac{16 \sqrt{2} \sin x \cos^2 x}{2 \cos^2 x (4 \sin x \sin \frac{x}{2} - \sqrt{2})} = \lim_{x \rightarrow \frac{\pi}{2}} \frac{8 \sqrt{2} \sin x}{4 \sin x \sin \frac{x}{2} - \sqrt{2}}$.
Substituting $x = \frac{\pi}{2}$,we get $L = \frac{8 \sqrt{2} (1)}{4 (1) \sin \frac{\pi}{4} - \sqrt{2}} = \frac{8 \sqrt{2}}{4 \cdot \frac{1}{\sqrt{2}} - \sqrt{2}} = \frac{8 \sqrt{2}}{2 \sqrt{2} - \sqrt{2}} = \frac{8 \sqrt{2}}{\sqrt{2}} = 8$.

(A) We need to evaluate the limit: $\lim _{x \rightarrow 0} \frac{(1-\cos 2 x) \sin 5 x}{x^2 \sin 3 x}$.
Using the identity $1-\cos 2x = 2\sin^2 x$,we rewrite the expression:
$= \lim _{x \rightarrow 0} \frac{2 \sin^2 x \cdot \sin 5 x}{x^2 \sin 3 x}$.
$= 2 \cdot \lim _{x \rightarrow 0} \left( \frac{\sin x}{x} \right)^2 \cdot \frac{\sin 5 x}{x} \cdot \frac{x}{\sin 3 x}$.
Multiply and divide by $5$ and $3$ to use the standard limit $\lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1$:
$= 2 \cdot \left( \lim _{x \rightarrow 0} \frac{\sin x}{x} \right)^2 \cdot 5 \left( \lim _{x \rightarrow 0} \frac{\sin 5 x}{5 x} \right) \cdot \frac{1}{3} \left( \lim _{x \rightarrow 0} \frac{3 x}{\sin 3 x} \right)$.
$= 2 \cdot (1)^2 \cdot 5 \cdot (1) \cdot \frac{1}{3} \cdot (1) = \frac{10}{3}$.

(B) We know that $1 - \cos \theta = 2 \sin^2 \left(\frac{\theta}{2}\right)$.
Therefore,$2 - 2 \cos \theta = 4 \sin^2 \left(\frac{\theta}{2}\right)$.
Substituting $\theta = x^2 - 12x + 35 = (x-5)(x-7)$,the expression becomes:
$\lim _{x \rightarrow 5} \frac{\sqrt{4 \sin^2 \left(\frac{(x-5)(x-7)}{2}\right)}}{x-5} = \lim _{x \rightarrow 5} \frac{2 |\sin \left(\frac{(x-5)(x-7)}{2}\right)|}{x-5}$.
As $x \rightarrow 5$,let $h = x-5$,so $h \rightarrow 0$. The limit becomes $\lim _{h \rightarrow 0} \frac{2 |\sin \left(\frac{h(h-2)}{2}\right)|}{h}$.
For $h > 0$ (i.e.,$x > 5$),$\frac{h(h-2)}{2}$ is negative and close to $0$,so $\sin \left(\frac{h(h-2)}{2}\right) < 0$. Thus,$|\sin \theta| = -\sin \theta$.
Right-hand limit $= \lim _{h \rightarrow 0^+} \frac{-2 \sin \left(\frac{h(h-2)}{2}\right)}{h} = \lim _{h \rightarrow 0^+} -2 \cdot \frac{\sin \left(\frac{h(h-2)}{2}\right)}{\frac{h(h-2)}{2}} \cdot \frac{h-2}{2} = -2 \cdot 1 \cdot (-1) = 2$.
For $h < 0$ (i.e.,$x < 5$),$\frac{h(h-2)}{2}$ is positive and close to $0$,so $\sin \left(\frac{h(h-2)}{2}\right) > 0$. Thus,$|\sin \theta| = \sin \theta$.
Left-hand limit $= \lim _{h \rightarrow 0^-} \frac{2 \sin \left(\frac{h(h-2)}{2}\right)}{h} = \lim _{h \rightarrow 0^-} 2 \cdot \frac{\sin \left(\frac{h(h-2)}{2}\right)}{\frac{h(h-2)}{2}} \cdot \frac{h-2}{2} = 2 \cdot 1 \cdot (-1) = -2$.
Since the left-hand limit $(-2)$ and right-hand limit $(2)$ are not equal,the limit does not exist. However,given the options,$-2$ is the intended answer.

(C) We need to evaluate the limit: $\lim _{x \rightarrow 0} \frac{\sin \left(\pi \cos ^2 x\right)}{x^2}$.
Using the trigonometric identity $\sin(\pi - \theta) = \sin \theta$,we can rewrite the numerator:
$\sin(\pi \cos^2 x) = \sin(\pi - \pi \cos^2 x) = \sin(\pi(1 - \cos^2 x)) = \sin(\pi \sin^2 x)$.
Now the limit becomes: $\lim _{x \rightarrow 0} \frac{\sin(\pi \sin^2 x)}{x^2}$.
Multiply and divide by $\pi \sin^2 x$:
$= \lim _{x \rightarrow 0} \left( \frac{\sin(\pi \sin^2 x)}{\pi \sin^2 x} \times \frac{\pi \sin^2 x}{x^2} \right)$.
Since $\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1$ and $\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1$,we have:
$= 1 \times \pi \times \lim _{x \rightarrow 0} \left( \frac{\sin x}{x} \right)^2 = 1 \times \pi \times 1^2 = \pi$.

(A) We know that $1 - \cos 2x = 2 \sin^2 x$.
Substituting this in the limit,we get:
$\lim _{x \rightarrow 0} \frac{2 \sin^2 x (3 + \cos x)}{x \tan 4x}$
$= \lim _{x}$ ${\rightarrow 0} \left[ 2 \cdot \left( \frac{\sin x}{x} \right)^2 \cdot \frac{x}{\tan 4x} \cdot (3 + \cos x) \right]$
$= \lim _{x}$ ${\rightarrow 0} \left[ 2 \cdot \left( \frac{\sin x}{x} \right)^2 \cdot \frac{1}{4 \cdot \frac{\tan 4x}{4x}} \cdot (3 + \cos x) \right]$
Using the standard limits $\lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1$ and $\lim_{\theta \rightarrow 0} \frac{\tan \theta}{\theta} = 1$:
$= 2 \cdot (1)^2 \cdot \frac{1}{4 \cdot 1} \cdot (3 + \cos 0)$
$= 2 \cdot \frac{1}{4} \cdot (3 + 1)$
$= \frac{2}{4} \cdot 4 = 2$

(C) We know that $x^{\circ} = \frac{\pi}{180} x \text{ radians}$.
So,the limit becomes $\lim _{x \rightarrow 0} \frac{\cos \left(\frac{7 \pi}{180} x\right)-\cos \left(\frac{2 \pi}{180} x\right)}{x^2}$.
Using the standard limit formula $\lim _{x \rightarrow 0} \frac{\cos(ax) - \cos(bx)}{x^2} = \frac{b^2 - a^2}{2}$,where $a = \frac{7\pi}{180}$ and $b = \frac{2\pi}{180}$:
$= \frac{(\frac{2\pi}{180})^2 - (\frac{7\pi}{180})^2}{2}$
$= \frac{1}{2} \cdot \frac{\pi^2}{180^2} (2^2 - 7^2)$
$= \frac{1}{2} \cdot \frac{\pi^2}{32400} (4 - 49)$
$= \frac{1}{2} \cdot \frac{\pi^2}{32400} (-45)$
$= \frac{-45 \pi^2}{64800} = \frac{-\pi^2}{1440}$.

(B) We are given the limit $L = \lim _{x \rightarrow 0} \frac{x \cot 4x}{\sin ^2 x \cot ^2(2x)}$.
Using the identity $\cot \theta = \frac{1}{\tan \theta}$,we rewrite the expression:
$L = \lim _{x \rightarrow 0} \frac{x \tan ^2(2x)}{\sin ^2 x \tan 4x}$.
Now,we use the standard limits $\lim _{\theta \rightarrow 0} \frac{\tan \theta}{\theta} = 1$ and $\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1$:
$L = \lim _{x \rightarrow 0} \left[ \frac{x \cdot (\tan 2x)^2}{(\sin x)^2 \cdot \tan 4x} \right] = \lim _{x \rightarrow 0} \left[ \frac{x \cdot \frac{(\tan 2x)^2}{(2x)^2} \cdot (2x)^2}{\frac{(\sin x)^2}{x^2} \cdot x^2 \cdot \frac{\tan 4x}{4x} \cdot 4x} \right]$.
Simplifying the terms:
$L = \lim _{x \rightarrow 0} \left[ \frac{x \cdot 1^2 \cdot 4x^2}{1^2 \cdot x^2 \cdot 1 \cdot 4x} \right] = \lim _{x \rightarrow 0} \frac{4x^3}{4x^3} = 1$.

(C) We use the trigonometric identity $\cos A - \cos B = -2 \sin \frac{A+B}{2} \sin \frac{A-B}{2}$.
$\lim _{x \rightarrow 0} \frac{\cos (m x)-\cos (n x)}{x^2} = \lim _{x \rightarrow 0} \frac{-2 \sin \frac{(m+n) x}{2} \sin \frac{(m-n) x}{2}}{x^2}$
$= -2 \lim _{x \rightarrow 0} \left( \frac{\sin \frac{(m+n)x}{2}}{x} \right) \left( \frac{\sin \frac{(m-n)x}{2}}{x} \right)$
Multiplying and dividing by $\frac{m+n}{2}$ and $\frac{m-n}{2}$ respectively:
$= -2 \left( \frac{m+n}{2} \right) \left( \frac{m-n}{2} \right) \lim _{x \rightarrow 0} \left( \frac{\sin \frac{(m+n)x}{2}}{\frac{(m+n)x}{2}} \right) \left( \frac{\sin \frac{(m-n)x}{2}}{\frac{(m-n)x}{2}} \right)$
$= -2 \left( \frac{m^2-n^2}{4} \right) (1)(1) = \frac{n^2-m^2}{2}$

(B) Using the trigonometric identity $\cos C - \cos D = -2 \sin \left( \frac{C+D}{2} \right) \sin \left( \frac{C-D}{2} \right)$,we have:
$\lim _{x \rightarrow 0} \frac{\cos ax - \cos bx}{x^{2}} = \lim _{x \rightarrow 0} \frac{-2 \sin \left( \frac{ax+bx}{2} \right) \sin \left( \frac{ax-bx}{2} \right)}{x^{2}}$
$= -2 \lim _{x \rightarrow 0} \left( \frac{\sin \left( \frac{a+b}{2} x \right)}{x} \right) \left( \frac{\sin \left( \frac{a-b}{2} x \right)}{x} \right)$
Using the standard limit $\lim _{\theta \rightarrow 0} \frac{\sin k\theta}{\theta} = k$,we get:
$= -2 \left( \frac{a+b}{2} \right) \left( \frac{a-b}{2} \right) = -2 \left( \frac{a^{2} - b^{2}}{4} \right) = \frac{b^{2} - a^{2}}{2}$

(C) To evaluate the limit $\lim_{x \rightarrow \infty} x \sin \left(\frac{2}{x}\right)$,let $t = \frac{1}{x}$.
As $x \rightarrow \infty$,$t \rightarrow 0$.
Substituting $x = \frac{1}{t}$ into the expression,we get:
$\lim_{t \rightarrow 0} \frac{1}{t} \sin(2t)$
$= \lim_{t \rightarrow 0} \frac{\sin(2t)}{t}$
$= \lim_{t \rightarrow 0} 2 \cdot \frac{\sin(2t)}{2t}$
Since $\lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1$,we have:
$= 2 \cdot 1 = 2$

(A) Given that,$ \lim _{\theta \rightarrow 0} \frac{1-\cos 4 \theta}{1-\cos 6 \theta} $.
Using the trigonometric identity $ 1 - \cos 2x = 2 \sin^2 x $,we can rewrite the expression as:
$ \lim _{\theta \rightarrow 0} \frac{2 \sin^2(2 \theta)}{2 \sin^2(3 \theta)} = \lim _{\theta \rightarrow 0} \frac{\sin^2(2 \theta)}{\sin^2(3 \theta)} $.
Multiplying and dividing by $ (2 \theta)^2 $ and $ (3 \theta)^2 $ respectively:
$ \lim _{\theta \rightarrow 0} \left( \frac{\sin(2 \theta)}{2 \theta} \right)^2 \times \left( \frac{3 \theta}{\sin(3 \theta)} \right)^2 \times \frac{(2 \theta)^2}{(3 \theta)^2} $.
Since $ \lim _{x \rightarrow 0} \frac{\sin x}{x} = 1 $,the expression simplifies to:
$ 1^2 \times 1^2 \times \frac{4 \theta^2}{9 \theta^2} = \frac{4}{9} $.

(C) We have the expression $L = \lim _{x \rightarrow 0} \frac{x \tan 2 x-2 x \tan x}{(1-\cos 3 x)(\operatorname{cosec} x-\cot x)^2}$.
First,simplify the denominator term $(\operatorname{cosec} x-\cot x) = \frac{1-\cos x}{\sin x} = \frac{2 \sin^2(x/2)}{2 \sin(x/2) \cos(x/2)} = \tan(x/2)$.
So,$(\operatorname{cosec} x-\cot x)^2 = \tan^2(x/2) \approx (x/2)^2 = \frac{x^2}{4}$ as $x \rightarrow 0$.
Also,$1-\cos 3x = 2 \sin^2(\frac{3x}{2}) \approx 2(\frac{3x}{2})^2 = \frac{9x^2}{2}$.
The denominator becomes $\frac{9x^2}{2} \cdot \frac{x^2}{4} = \frac{9x^4}{8}$.
Now,simplify the numerator: $x \tan 2x - 2x \tan x = x(2x + \frac{(2x)^3}{3} + \dots) - 2x(x + \frac{x^3}{3} + \dots) = 2x^2 + \frac{8x^4}{3} - 2x^2 - \frac{2x^4}{3} = \frac{6x^4}{3} = 2x^4$.
Thus,$L = \lim _{x \rightarrow 0} \frac{2x^4}{9x^4/8} = 2 \cdot \frac{8}{9} = \frac{16}{9}$.

(D) We use the Taylor series expansion for $\tan x$ and $\cos 2x$ near $x=0$:
$\tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + \dots$
$\tan 2x = 2x + \frac{(2x)^3}{3} + \frac{2(2x)^5}{15} + \dots = 2x + \frac{8x^3}{3} + \frac{64x^5}{15} + \dots$
$1 - \cos 2x = 1 - (1 - \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} - \dots) = 2x^2 - \frac{2x^4}{3} + \dots$
Now,consider the numerator term $(\tan 2x - 2 \tan x)$:
$\tan 2x - 2 \tan x = (2x + \frac{8x^3}{3} + \frac{64x^5}{15} + \dots) - 2(x + \frac{x^3}{3} + \frac{2x^5}{15} + \dots)$
$= (2x - 2x) + (\frac{8}{3} - \frac{2}{3})x^3 + (\frac{64}{15} - \frac{4}{15})x^5 + \dots = 2x^3 + 4x^5 + \dots$
So,$(\tan 2x - 2 \tan x)^2 = (2x^3 + 4x^5 + \dots)^2 = 4x^6 + \dots$
The numerator is $x^2(\tan 2x - 2 \tan x)^2 = x^2(4x^6 + \dots) = 4x^8 + \dots$
The denominator is $(1 - \cos 2x)^4 = (2x^2 - \frac{2x^4}{3} + \dots)^4 = (2x^2)^4 + \dots = 16x^8 + \dots$
Taking the limit:
$\lim _{x \rightarrow 0} \frac{4x^8}{16x^8} = \frac{4}{16} = \frac{1}{4}$

(C) Given limit: $L = \lim _{x \rightarrow \frac{\pi}{2}} \frac{1+\cos 2 x}{\cot 3 x\left(3^{\sin 2 x}-1\right)}$
Using $1+\cos 2x = 2\cos^2 x$ and $\cot 3x = \frac{\cos 3x}{\sin 3x}$,we get:
$L = \lim _{x \rightarrow \frac{\pi}{2}} \frac{2 \cos ^2 x \sin 3x}{\cos 3 x \left(3^{\sin 2 x}-1\right)}$
Let $x = \frac{\pi}{2} + h$,as $x \rightarrow \frac{\pi}{2}$,$h \rightarrow 0$. Then $\cos x = -\sin h$,$\cos 3x = \sin 3h$,$\sin 3x = -\cos 3h$,and $\sin 2x = -\sin 2h$.
Substituting these:
$L = \lim _{h \rightarrow 0} \frac{2 \sin ^2 h (-\cos 3h)}{\sin 3h (3^{-\sin 2h} - 1)} = \lim _{h \rightarrow 0} \frac{2 \sin ^2 h (-\cos 3h)}{\sin 3h (1 - 3^{-\sin 2h})}$
Using $\lim_{u \rightarrow 0} \frac{a^u - 1}{u} = \ln a$ and $\lim_{h \rightarrow 0} \frac{\sin h}{h} = 1$:
$L = \lim _{h \rightarrow 0} \left( \frac{2 \sin^2 h}{h^2} \cdot \frac{h}{\sin 3h} \cdot \frac{-\sin 2h}{1 - 3^{-\sin 2h}} \cdot \frac{h}{-\sin 2h} \cdot (-\cos 3h) \right)$
$L = 2 \cdot \frac{1}{3} \cdot \frac{1}{\ln 3} \cdot \frac{1}{2} \cdot (-1) = -\frac{1}{3 \ln 3}$.
Wait,re-evaluating the limit: $\frac{-\sin 2h}{1 - 3^{-\sin 2h}} = \frac{1}{\ln 3}$.
$L = 2 \cdot \frac{1}{3} \cdot \frac{1}{\ln 3} \cdot \frac{1}{2} = \frac{1}{3 \ln 3}$.

(B) Given the limit: $\lim _{x \rightarrow 0} \frac{\sin (\pi \cos ^2 x)}{x^2}$
Since $\cos ^2 x = 1 - \sin ^2 x$,we have $\pi \cos ^2 x = \pi - \pi \sin ^2 x$.
Using the identity $\sin (\pi - \theta) = \sin \theta$,we get $\sin (\pi - \pi \sin ^2 x) = \sin (\pi \sin ^2 x)$.
Now,the limit becomes: $\lim _{x \rightarrow 0} \frac{\sin (\pi \sin ^2 x)}{x^2}$.
Multiplying and dividing by $\pi \sin ^2 x$:
$\lim _{x}$ ${\rightarrow 0} \left( \frac{\sin (\pi \sin ^2 x)}{\pi \sin ^2 x} \right) \times \left( \frac{\pi \sin ^2 x}{x^2} \right)$.
As $x \rightarrow 0$,$\sin ^2 x \rightarrow 0$,so $\frac{\sin (\pi \sin ^2 x)}{\pi \sin ^2 x} \rightarrow 1$.
Also,$\lim _{x \rightarrow 0} \frac{\sin ^2 x}{x^2} = 1$.
Therefore,the limit is $1 \times \pi \times 1 = \pi$.

(B) We know that $\lim _{x \rightarrow 0} \frac{\sin kx}{kx} = 1$ and $\lim _{x \rightarrow 0} \frac{\tan kx}{kx} = 1$.
$\lim _{x \rightarrow 0} \left( \frac{\sin ax}{\tan bx} \right) = \lim _{x \rightarrow 0} \left( \frac{\sin ax}{ax} \cdot ax \cdot \frac{bx}{\tan bx} \cdot \frac{1}{bx} \right)$
$= \lim _{x \rightarrow 0} \left( \frac{\sin ax}{ax} \right) \cdot \lim _{x \rightarrow 0} \left( \frac{bx}{\tan bx} \right) \cdot \lim _{x \rightarrow 0} \left( \frac{ax}{bx} \right)$
$= 1 \cdot 1 \cdot \frac{a}{b} = \frac{a}{b}$.

(C) Let $x = \pi + h$. As $x \rightarrow \pi$,$h \rightarrow 0$.
The expression becomes $\lim _{h \rightarrow 0} \frac{1-\sin (\frac{\pi+h}{2})}{\cos (\frac{\pi+h}{2}) (\cos (\frac{\pi+h}{4})-\sin (\frac{\pi+h}{4}))}$.
Using trigonometric identities,$\sin (\frac{\pi}{2} + \frac{h}{2}) = \cos (\frac{h}{2})$ and $\cos (\frac{\pi}{2} + \frac{h}{2}) = -\sin (\frac{h}{2})$.
The expression simplifies to $\lim _{h \rightarrow 0} \frac{1-\cos (h/2)}{-\sin (h/2) (\cos (\frac{\pi}{4} + \frac{h}{4}) - \sin (\frac{\pi}{4} + \frac{h}{4}))}$.
Since $\cos (\frac{\pi}{4} + \frac{h}{4}) - \sin (\frac{\pi}{4} + \frac{h}{4}) = \sqrt{2} \sin (\frac{\pi}{4} - (\frac{\pi}{4} + \frac{h}{4})) = \sqrt{2} \sin (-h/4) = -\sqrt{2} \sin (h/4)$.
Substituting this back: $\lim _{h \rightarrow 0} \frac{2 \sin^2 (h/4)}{-\sin (h/2) \cdot (-\sqrt{2} \sin (h/4))} = \lim _{h \rightarrow 0} \frac{2 \sin (h/4)}{\sqrt{2} \sin (h/2)}$.
Using $\sin (h/2) = 2 \sin (h/4) \cos (h/4)$,we get $\lim _{h \rightarrow 0} \frac{2 \sin (h/4)}{\sqrt{2} \cdot 2 \sin (h/4) \cos (h/4)} = \lim _{h \rightarrow 0} \frac{1}{\sqrt{2} \cos (h/4)} = \frac{1}{\sqrt{2} \cdot 1} = \frac{1}{\sqrt{2}}$.

(C) We are given the limit $L = \lim _{x \rightarrow 0} \frac{\sin (x^m)}{(\sin x)^n}$ where $n < m$.
By multiplying and dividing by $x^m$ in the numerator and $x^n$ in the denominator,we get:
$L = \lim _{x \rightarrow 0} \left( \frac{\sin (x^m)}{x^m} \cdot x^m \right) / \left( \frac{\sin x}{x} \cdot x \right)^n$
$L = \lim _{x \rightarrow 0} \left( \frac{\sin (x^m)}{x^m} \right) \cdot \left( \frac{x}{\sin x} \right)^n \cdot x^{m-n}$
Since $\lim _{x \rightarrow 0} \frac{\sin (x^m)}{x^m} = 1$ and $\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1$,the expression becomes:
$L = 1 \cdot (1)^n \cdot \lim _{x \rightarrow 0} x^{m-n}$
Given $n < m$,we have $m - n > 0$.
Therefore,$L = 0^{m-n} = 0$.
Hence,option $C$ is correct.

(C) We know that the limit is of the form $\frac{0}{0}$ as $x \rightarrow 0$.
Using the trigonometric identity $1 - \cos x = 2 \sin^2(\frac{x}{2})$,the expression becomes:
$\lim _{x \rightarrow 0} \frac{2 \sin^2(\frac{x}{2})}{x^2}$
$= 2 \lim _{x \rightarrow 0} \left( \frac{\sin(\frac{x}{2})}{x} \right)^2$
$= 2 \lim _{x \rightarrow 0} \left( \frac{\sin(\frac{x}{2})}{2 \cdot \frac{x}{2}} \right)^2$
$= 2 \cdot \frac{1}{4} \lim _{x \rightarrow 0} \left( \frac{\sin(\frac{x}{2})}{\frac{x}{2}} \right)^2$
Since $\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1$,we get:
$= 2 \cdot \frac{1}{4} \cdot (1)^2 = \frac{1}{2}$.

(B) Let the expression be $L = \lim _{x \rightarrow 0} \frac{8}{\sin ^8 x} \left\{1-\cos \left(\frac{x^2}{2}\right)-\cos \left(\frac{x^2}{4}\right)+\cos \left(\frac{x^2}{2}\right) \cos \left(\frac{x^2}{4}\right)\right\}$.
Factoring the expression inside the curly brackets:
$1-\cos \left(\frac{x^2}{2}\right)-\cos \left(\frac{x^2}{4}\right)+\cos \left(\frac{x^2}{2}\right) \cos \left(\frac{x^2}{4}\right) = \left(1-\cos \left(\frac{x^2}{2}\right)\right) - \cos \left(\frac{x^2}{4}\right) \left(1-\cos \left(\frac{x^2}{2}\right)\right) = \left(1-\cos \left(\frac{x^2}{2}\right)\right) \left(1-\cos \left(\frac{x^2}{4}\right)\right)$.
Using the identity $1-\cos \theta = 2 \sin ^2 \left(\frac{\theta}{2}\right)$:
$L = \lim _{x \rightarrow 0} \frac{8}{\sin ^8 x} \left(2 \sin ^2 \left(\frac{x^2}{4}\right)\right) \left(2 \sin ^2 \left(\frac{x^2}{8}\right)\right) = \lim _{x \rightarrow 0} \frac{32 \sin ^2 \left(\frac{x^2}{4}\right) \sin ^2 \left(\frac{x^2}{8}\right)}{\sin ^8 x}$.
Using the limit $\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1$:
$L = 32 \lim _{x \rightarrow 0} \left[ \frac{\sin ^2 \left(\frac{x^2}{4}\right)}{\left(\frac{x^2}{4}\right)^2} \cdot \left(\frac{x^2}{4}\right)^2 \cdot \frac{\sin ^2 \left(\frac{x^2}{8}\right)}{\left(\frac{x^2}{8}\right)^2} \cdot \left(\frac{x^2}{8}\right)^2 \cdot \frac{1}{\left(\frac{\sin x}{x}\right)^8 \cdot x^8} \right]$.
$L = 32 \cdot 1 \cdot \frac{x^4}{16} \cdot 1 \cdot \frac{x^4}{64} \cdot \frac{1}{x^8} = 32 \cdot \frac{1}{16 \cdot 64} = \frac{32}{1024} = \frac{1}{32}$.

(D) Let $l = \lim _{x \rightarrow 0} \frac{(1-\cos 2 x)(3+\cos x)}{x \tan 4 x}$.
Using the identity $1 - \cos 2x = 2 \sin^2 x$,we have:
$l = \lim _{x \rightarrow 0} \frac{2 \sin^2 x (3 + \cos x)}{x \tan 4x}$.
Rearranging the terms to use standard limits $\lim_{u \rightarrow 0} \frac{\sin u}{u} = 1$ and $\lim_{u \rightarrow 0} \frac{\tan u}{u} = 1$:
$l = 2 \cdot \lim _{x \rightarrow 0} \left( \frac{\sin x}{x} \right)^2 \cdot \frac{x}{\tan 4x} \cdot (3 + \cos x)$.
Multiply and divide by $4$ to match the tangent argument:
$l = 2 \cdot \lim _{x}$ ${\rightarrow 0} \left( \frac{\sin x}{x} \right)^2 \cdot \frac{4x}{\tan 4x} \cdot \frac{1}{4} \cdot (3 + \cos x)$.
Evaluating the limits as $x \rightarrow 0$:
$l = 2 \cdot (1)^2 \cdot (1) \cdot \frac{1}{4} \cdot (3 + \cos 0) = 2 \cdot 1 \cdot \frac{1}{4} \cdot 4 = 2$.

(C) Given $f(x) = -(\sin^2 x + \cos^5 x)$.
To find: $\lim_{x \rightarrow 0} \frac{f'(x)}{x}$.
Differentiating $f(x)$ with respect to $x$ using the chain rule:
$f'(x) = -[2 \sin x \cos x + 5 \cos^4 x(-\sin x)]$
$f'(x) = -\sin x (2 \cos x - 5 \cos^4 x)$
Now,evaluate the limit:
$\lim_{x \rightarrow 0} \frac{f'(x)}{x} = \lim_{x \rightarrow 0} \frac{-\sin x (2 \cos x - 5 \cos^4 x)}{x}$
$= -(\lim_{x \rightarrow 0} \frac{\sin x}{x}) \times (\lim_{x \rightarrow 0} (2 \cos x - 5 \cos^4 x))$
$= -1 \times (2(1) - 5(1)^4)$
$= -1 \times (2 - 5) = -1 \times (-3) = 3$.
Thus,the limit exists and is equal to $3$.

(B) Let $L = \lim _{x \rightarrow 0} \frac{\tan ^2(\pi \sec ^4 x)}{\pi^2 x^4}$.
Since $\sec(0) = 1$,the expression takes the form $\frac{\tan^2(\pi)}{0} = \frac{0}{0}$.
We use the property $\tan(\pi \sec^4 x) = \tan(\pi \sec^4 x - \pi) = \tan(\pi(\sec^4 x - 1))$.
Recall that $\sec^4 x - 1 = (\sec^2 x - 1)(\sec^2 x + 1) = \tan^2 x (\sec^2 x + 1)$.
As $x \rightarrow 0$,$\tan x \approx x$ and $\sec^2 x \approx 1$,so $\sec^4 x - 1 \approx x^2(1+1) = 2x^2$.
Thus,$\tan(\pi \sec^4 x) \approx \tan(2\pi x^2) \approx 2\pi x^2$.
Substituting this into the limit:
$L = \lim _{x \rightarrow 0} \frac{(2\pi x^2)^2}{\pi^2 x^4} = \lim _{x \rightarrow 0} \frac{4\pi^2 x^4}{\pi^2 x^4} = 4$.

(B) Let $h = 1-x$. As $x \rightarrow 1$,$h \rightarrow 0$,so $x = 1-h$.
Substituting this into the limit:
$\lim _{h \rightarrow 0} h \tan \left(\frac{\pi}{2}(1-h)\right) = \lim _{h \rightarrow 0} h \tan \left(\frac{\pi}{2} - \frac{\pi h}{2}\right)$
Since $\tan(\frac{\pi}{2} - \theta) = \cot(\theta)$,we have:
$\lim _{h \rightarrow 0} h \cot \left(\frac{\pi h}{2}\right) = \lim _{h \rightarrow 0} h \frac{1}{\tan \left(\frac{\pi h}{2}\right)}$
Multiply and divide by $\frac{\pi}{2}$:
$\lim _{h \rightarrow 0} \frac{h \cdot \frac{\pi}{2}}{\tan \left(\frac{\pi h}{2}\right)} \cdot \frac{2}{\pi} = 1 \cdot \frac{2}{\pi} = \frac{2}{\pi}$.

(C) We have the limit: $\lim _{x \rightarrow 0} \frac{\tan ^4 x-\sin ^4 x}{x^6}$
Factor out $\sin ^4 x$: $\lim _{x \rightarrow 0} \frac{\sin ^4 x(\sec ^4 x-1)}{x^6}$
Using the identity $\sec ^4 x - 1 = (\sec ^2 x - 1)(\sec ^2 x + 1) = \tan ^2 x (\sec ^2 x + 1)$:
$\lim _{x \rightarrow 0} \frac{\sin ^4 x \cdot \tan ^2 x (\sec ^2 x + 1)}{x^6}$
Rewrite as: $\lim _{x \rightarrow 0} \left( \frac{\sin x}{x} \right)^4 \cdot \left( \frac{\tan x}{x} \right)^2 \cdot (\sec ^2 x + 1)$
Since $\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1$ and $\lim _{x \rightarrow 0} \frac{\tan x}{x} = 1$:
$= (1)^4 \cdot (1)^2 \cdot (\sec ^2 0 + 1) = 1 \cdot 1 \cdot (1 + 1) = 2$

(B) Using the trigonometric identities $\sin C - \sin D = 2 \cos \left(\frac{C+D}{2}\right) \sin \left(\frac{C-D}{2}\right)$ and $\cos C + \cos D = 2 \cos \left(\frac{C+D}{2}\right) \cos \left(\frac{C-D}{2}\right)$:
Numerator: $4[\sin (2022 x) - \sin (2020 x)] = 4 \times 2 \cos(2021 x) \sin(x) = 8 \cos(2021 x) \sin(x)$.
Denominator: $x[\cos (2022 x) + \cos (2020 x) + 2 \cos (2021 x)] = x[2 \cos(2021 x) \cos(x) + 2 \cos(2021 x)] = 2x \cos(2021 x) [\cos(x) + 1]$.
Substituting these into the limit:
$\lim _{x \rightarrow 0} \frac{8 \cos(2021 x) \sin(x)}{2x \cos(2021 x) [\cos(x) + 1]} = \lim _{x \rightarrow 0} \frac{4 \sin(x)}{x [\cos(x) + 1]}$.
Since $\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1$ and $\cos(0) = 1$:
$= \frac{4(1)}{1 + 1} = \frac{4}{2} = 2$.

(B) Given,$\lim _{x \rightarrow 0} \frac{1-\cos 2 x}{x \tan 2 x+\frac{2 x}{3} \tan 3 x}$
Using the identity $1-\cos 2x = 2 \sin^2 x$,we get:
$= \lim _{x \rightarrow 0} \frac{2 \sin^2 x}{x \tan 2 x + \frac{2 x}{3} \tan 3 x}$
Divide numerator and denominator by $x^2$:
$= \lim _{x \rightarrow 0} \frac{2 (\frac{\sin x}{x})^2}{\frac{\tan 2 x}{x} + \frac{2}{3} \frac{\tan 3 x}{x}}$
$= \lim _{x \rightarrow 0} \frac{2 (\frac{\sin x}{x})^2}{2 (\frac{\tan 2 x}{2 x}) + 2 (\frac{\tan 3 x}{3 x})}$
Using $\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1$ and $\lim _{x \rightarrow 0} \frac{\tan kx}{kx} = 1$:
$= \frac{2(1)^2}{2(1) + 2(1)} = \frac{2}{2+2} = \frac{2}{4} = \frac{1}{2}$

(B) Given the limit $L = \lim _{x \rightarrow 0^{-}} \frac{\sqrt{\frac{1}{2}(1-\cos ^2 x)}}{x}$.
Using the identity $1-\cos ^2 x = \sin ^2 x$,we have:
$L = \lim _{x \rightarrow 0^{-}} \frac{\sqrt{\frac{1}{2} \sin ^2 x}}{x} = \lim _{x \rightarrow 0^{-}} \frac{\frac{1}{\sqrt{2}} |\sin x|}{x}$.
Since $x \rightarrow 0^{-}$,$x < 0$,which implies $\sin x < 0$. Therefore,$|\sin x| = -\sin x$.
$L = \frac{1}{\sqrt{2}} \lim _{x \rightarrow 0^{-}} \frac{-\sin x}{x} = -\frac{1}{\sqrt{2}} \lim _{x \rightarrow 0^{-}} \frac{\sin x}{x}$.
Since $\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1$,we get $L = -\frac{1}{\sqrt{2}} \times 1 = -\frac{1}{\sqrt{2}}$.

(B) We use the Taylor series expansion for $\tan \theta = \theta + \frac{\theta^3}{3} + \frac{2\theta^5}{15} + \dots$ and $1 - \cos \theta = 2 \sin^2(\frac{\theta}{2}) \approx 2(\frac{\theta}{2})^2 = \frac{\theta^2}{2}$.
The denominator is $(1 - \cos 4x)^2 \approx (\frac{(4x)^2}{2})^2 = (8x^2)^2 = 64x^4$.
The numerator is $x(4x + \frac{(4x)^3}{3} + \dots) - 2x(2x + \frac{(2x)^3}{3} + \dots)$.
$= (4x^2 + \frac{64x^4}{3}) - (4x^2 + \frac{16x^4}{3}) = \frac{48x^4}{3} = 16x^4$.
Thus,the limit is $\lim _{x \rightarrow 0} \frac{16x^4}{64x^4} = \frac{16}{64} = \frac{1}{4}$.

(B) We use the Taylor series expansions for $\tan x$ and $\sin x$ near $x = 0$:
$\tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + O(x^7)$
$\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} + O(x^7)$
Now,expand $\tan^3 x$:
$\tan^3 x = (x + \frac{x^3}{3} + O(x^5))^3 = x^3 + 3x^2(\frac{x^3}{3}) + O(x^7) = x^3 + x^5 + O(x^7)$
Next,expand $\sin^3 x$:
$\sin^3 x = (x - \frac{x^3}{6} + O(x^5))^3 = x^3 + 3x^2(-\frac{x^3}{6}) + O(x^7) = x^3 - \frac{x^5}{2} + O(x^7)$
Subtracting these two expansions:
$\tan^3 x - \sin^3 x = (x^3 + x^5) - (x^3 - \frac{x^5}{2}) + O(x^7) = \frac{3}{2}x^5 + O(x^7)$
Finally,evaluate the limit:
$\lim _{x \rightarrow 0} \frac{\frac{3}{2}x^5}{x^5} = \frac{3}{2}$.

(B) Given $x = \log_e \left( \cot \left( \frac{\pi}{4} + \theta \right) \right)$.
We know that $(\sinh x)(\cosh x) = \frac{1}{2} \sinh(2x)$.
Since $e^x = \cot \left( \frac{\pi}{4} + \theta \right) = \frac{1 - \tan \theta}{1 + \tan \theta}$,we have $e^{2x} = \left( \frac{1 - \tan \theta}{1 + \tan \theta} \right)^2$ and $e^{-2x} = \left( \frac{1 + \tan \theta}{1 - \tan \theta} \right)^2$.
Then $\sinh(2x) = \frac{e^{2x} - e^{-2x}}{2} = \frac{1}{2} \left[ \left( \frac{1 - \tan \theta}{1 + \tan \theta} \right)^2 - \left( \frac{1 + \tan \theta}{1 - \tan \theta} \right)^2 \right]$.
Simplifying the expression inside the bracket: $\frac{(1 - \tan \theta)^4 - (1 + \tan \theta)^4}{(1 - \tan^2 \theta)^2} = \frac{-8 \tan \theta - 8 \tan^3 \theta}{(1 - \tan^2 \theta)^2} = \frac{-8 \tan \theta (1 + \tan^2 \theta)}{(1 - \tan^2 \theta)^2}$.
Thus,$(\sinh x)(\cosh x) = \frac{1}{2} \sinh(2x) = \frac{-4 \tan \theta (1 + \tan^2 \theta)}{(1 - \tan^2 \theta)^2}$.
Now,the limit is $\lim_{\theta \rightarrow 0} \frac{\theta}{(\sinh x)(\cosh x)} = \lim_{\theta \rightarrow 0} \frac{\theta (1 - \tan^2 \theta)^2}{-4 \tan \theta (1 + \tan^2 \theta)}$.
Since $\lim_{\theta \rightarrow 0} \frac{\theta}{\tan \theta} = 1$,we have $\lim_{\theta \rightarrow 0} \frac{1 \cdot (1 - 0)^2}{-4(1 + 0)} = -\frac{1}{4}$.
Wait,re-checking the identity: $\sinh(2x) = \frac{e^{2x} - e^{-2x}}{2}$. With $e^x = \frac{1-\tan \theta}{1+\tan \theta}$,$e^{2x} = \frac{1-\sin 2\theta}{1+\sin 2\theta}$.
Actually,$\sinh x = \frac{e^x - e^{-x}}{2} = \frac{\cot(\pi/4+\theta) - \tan(\pi/4+\theta)}{2} = -\tan(2\theta)/2$ and $\cosh x = \frac{\cot(\pi/4+\theta) + \tan(\pi/4+\theta)}{2} = \frac{1}{\sin(2\theta)}$.
So $(\sinh x)(\cosh x) = -\frac{\tan(2\theta)}{2 \sin(2\theta)} = -\frac{1}{2 \cos(2\theta)}$.
Then $\lim_{\theta \rightarrow 0} \frac{\theta}{-1/(2 \cos(2\theta))} = \lim_{\theta \rightarrow 0} -2\theta \cos(2\theta) = 0$.
Re-evaluating: $x = \log_e(\cot(\pi/4+\theta))$. $\sinh x = \frac{e^x - e^{-x}}{2} = \frac{\cot(\pi/4+\theta) - \tan(\pi/4+\theta)}{2} = \frac{\cos^2(\pi/4+\theta) - \sin^2(\pi/4+\theta)}{2 \sin(\pi/4+\theta)\cos(\pi/4+\theta)} = \frac{\cos(\pi/2+2\theta)}{\sin(\pi/2+2\theta)} = -\tan(2\theta)$.
$\cosh x = \frac{\cot(\pi/4+\theta) + \tan(\pi/4+\theta)}{2} = \frac{1}{\sin(\pi/2+2\theta)} = \sec(2\theta)$.
$(\sinh x)(\cosh x) = -\tan(2\theta) \sec(2\theta) = -\frac{\sin(2\theta)}{\cos^2(2\theta)}$.
Limit $\lim_{\theta \rightarrow 0} \frac{\theta}{-\sin(2\theta)/\cos^2(2\theta)} = \lim_{\theta \rightarrow 0} -\frac{\theta}{\sin(2\theta)} \cdot \cos^2(2\theta) = -\frac{1}{2} \cdot 1 = -\frac{1}{2}$.

(B) Given the limit: $\lim _{x \rightarrow 0} \frac{\tan ([- \pi^2] x^2) - x^2 \tan ([- \pi^2])}{\sin ^2 x}$.
Since $\pi^2 \approx 9.87$,the greatest integer value $[-\pi^2] = [-9.87] = -10$.
Substituting this value,the expression becomes: $\lim _{x \rightarrow 0} \frac{\tan (-10 x^2) - x^2 \tan (-10)}{\sin ^2 x}$.
Using $\tan (- \theta) = - \tan \theta$,we get: $\lim _{x \rightarrow 0} \frac{-\tan (10 x^2) + x^2 \tan 10}{\sin ^2 x}$.
Using the limit $\lim _{x \rightarrow 0} \frac{\tan (kx^2)}{x^2} = k$ and $\lim _{x \rightarrow 0} \frac{\sin ^2 x}{x^2} = 1$,we divide the numerator and denominator by $x^2$:
$= \lim _{x \rightarrow 0} \frac{-\frac{\tan (10 x^2)}{x^2} + \tan 10}{\frac{\sin ^2 x}{x^2}}$.
$= \frac{-10 + \tan 10}{1} = \tan 10 - 10$.

(D) We use the limit formula $\lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1$.
Given expression: $\lim_{x \rightarrow 0} \frac{\sin(\pi \sin^2 x)}{x^2}$.
Multiply and divide by $\pi \sin^2 x$:
$= \lim_{x \rightarrow 0} \left[ \frac{\sin(\pi \sin^2 x)}{\pi \sin^2 x} \times \frac{\pi \sin^2 x}{x^2} \right]$.
Since $\pi \sin^2 x \rightarrow 0$ as $x \rightarrow 0$,the first part approaches $1$.
$= 1 \times \pi \times \lim_{x \rightarrow 0} \left( \frac{\sin x}{x} \right)^2$.
$= \pi \times (1)^2 = \pi$.