Concept of limits, Evaluation of algebric limits Questions in English

(D) The limit of a continuous function $f(x) = x^n$ as $x$ approaches $a$ is simply the value of the function at $x = a$.
Since $f(x) = x^n$ is continuous for $x > 0$,we have:
$\lim_{x \rightarrow a} x^n = a^n$

(D) We are given the limit: $\lim _{n \rightarrow \infty} \frac{n !}{(n+1) !-n !}$
First,simplify the denominator by factoring out $n!$:
$(n+1)! - n! = n!(n+1) - n!(1) = n!(n+1-1) = n!(n)$
Now,substitute this back into the expression:
$\lim _{n \rightarrow \infty} \frac{n!}{n!(n)} = \lim _{n \rightarrow \infty} \frac{1}{n}$
As $n \rightarrow \infty$,the value of $\frac{1}{n}$ approaches $0$.
Therefore,the limit is $0$.

(D) For $l_1 = \lim_{x \rightarrow 2^{+}} (x^2 + [x])$: As $x \rightarrow 2^{+}$,$[x] = 2$. Therefore,$l_1 = 2^2 + 2 = 4 + 2 = 6$.
For $l_2 = \lim_{x \rightarrow 3^{-}} (2x - [x])$: As $x \rightarrow 3^{-}$,$[x] = 2$. Therefore,$l_2 = 2(3) - 2 = 6 - 2 = 4$.
For $l_3 = \lim_{x \rightarrow \frac{\pi}{2}} \left( \frac{\cos x}{x - \frac{\pi}{2}} \right)$: Let $y = x - \frac{\pi}{2}$. As $x \rightarrow \frac{\pi}{2}$,$y \rightarrow 0$. Then $x = y + \frac{\pi}{2}$.
$l_3 = \lim_{y \rightarrow 0} \frac{\cos(y + \frac{\pi}{2})}{y} = \lim_{y \rightarrow 0} \frac{-\sin y}{y} = -1$.
Comparing the values,we have $l_3 = -1$,$l_2 = 4$,and $l_1 = 6$.
Thus,$l_3 < l_2 < l_1$.

(A) Let $L = \lim _{x \rightarrow 0} \frac{\sqrt{2}-\sqrt{1+\cos x}}{\sqrt{15+\cos 2x}-4}$.
Using the identities $1+\cos x = 2\cos^2(x/2)$ and $\cos 2x = 2\cos^2 x - 1$,we have:
$L = \lim _{x \rightarrow 0} \frac{\sqrt{2}-\sqrt{2}\cos(x/2)}{\sqrt{14+2\cos^2 x}-4}$.
Rationalizing the numerator and denominator:
$L = \lim _{x \rightarrow 0} \frac{\sqrt{2}(1-\cos(x/2))}{\sqrt{14+2\cos^2 x}-4} \times \frac{1+\cos(x/2)}{1+\cos(x/2)} \times \frac{\sqrt{14+2\cos^2 x}+4}{\sqrt{14+2\cos^2 x}+4}$.
$L = \lim _{x \rightarrow 0} \frac{\sqrt{2}\sin^2(x/2)}{1+\cos(x/2)} \times \frac{\sqrt{14+2\cos^2 x}+4}{14+2\cos^2 x - 16}$.
Since $14+2\cos^2 x - 16 = 2\cos^2 x - 2 = -2\sin^2 x = -8\sin^2(x/2)\cos^2(x/2)$:
$L = \lim _{x \rightarrow 0} \frac{\sqrt{2}\sin^2(x/2)}{1+\cos(x/2)} \times \frac{\sqrt{14+2\cos^2 x}+4}{-8\sin^2(x/2)\cos^2(x/2)}$.
$L = \frac{\sqrt{2}}{1+1} \times \frac{\sqrt{14+2}+4}{-8(1)} = \frac{\sqrt{2}}{2} \times \frac{8}{-8} = -\frac{\sqrt{2}}{2} = -\frac{1}{\sqrt{2}}$.

(C) Given the limit $L = \lim _{x \rightarrow-\infty} \frac{5 x^3-x^2 \sin 5 x}{x \cos 4 x+7|x|^3-4|x|+3}$.
Since $x \rightarrow -\infty$,we have $x < 0$,so $|x| = -x$ and $|x|^3 = -x^3$.
Substituting these into the expression:
$L = \lim _{x \rightarrow-\infty} \frac{5 x^3-x^2 \sin 5 x}{x \cos 4 x + 7(-x^3) - 4(-x) + 3} = \lim _{x \rightarrow-\infty} \frac{5 x^3-x^2 \sin 5 x}{x \cos 4 x - 7x^3 + 4x + 3}$.
Divide the numerator and denominator by $x^3$:
$L = \lim _{x \rightarrow-\infty} \frac{5 - \frac{\sin 5 x}{x}}{\frac{\cos 4 x}{x^2} - 7 + \frac{4}{x^2} + \frac{3}{x^3}}$.
As $x \rightarrow -\infty$,$\frac{\sin 5 x}{x} \rightarrow 0$,$\frac{\cos 4 x}{x^2} \rightarrow 0$,$\frac{4}{x^2} \rightarrow 0$,and $\frac{3}{x^3} \rightarrow 0$.
Therefore,$L = \frac{5 - 0}{0 - 7 + 0 + 0} = -\frac{5}{7}$.

(D) First,evaluate the limit for $k$:
$\lim _{x \rightarrow 0} \frac{\cos 2x - \cos 4x}{1 - \cos 2x} = \lim _{x \rightarrow 0} \frac{(1 - 2\sin^2 x) - (1 - 8\sin^2 x \cos^2 x)}{2\sin^2 x}$
$= \lim _{x \rightarrow 0} \frac{8\sin^2 x \cos^2 x - 2\sin^2 x}{2\sin^2 x} = \lim _{x \rightarrow 0} (4\cos^2 x - 1) = 4(1)^2 - 1 = 3$.
So,$k = 3$.
Now,evaluate the second limit with $k = 3$:
$\lim _{x \rightarrow 3} \frac{x^3 - 27}{x^4 - 81} = \lim _{x \rightarrow 3} \frac{(x - 3)(x^2 + 3x + 9)}{(x - 3)(x + 3)(x^2 + 9)} = \lim _{x \rightarrow 3} \frac{x^2 + 3x + 9}{(x + 3)(x^2 + 9)} = \frac{9 + 9 + 9}{(3 + 3)(9 + 9)} = \frac{27}{6 \times 18} = \frac{27}{108} = \frac{1}{4}$.

(C) We know that the sum of the squares of the first $n$ natural numbers is given by $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$.
Substituting this into the expression,we get:
$\lim _{n \rightarrow \infty} \frac{1}{n^3} \left( \frac{n(n+1)(2n+1)}{6} \right) x$
$= \lim _{n \rightarrow \infty} \frac{n(n+1)(2n+1)}{6n^3} x$
$= \lim _{n \rightarrow \infty} \frac{n^3(1 + \frac{1}{n})(2 + \frac{1}{n})}{6n^3} x$
$= \frac{1 \times 2}{6} x = \frac{2}{6} x = \frac{x}{3}$.

(C) To evaluate the limit $\lim _{x \rightarrow \infty} \frac{3 x+4 \cos ^2 x}{\sqrt{x^2-5 \sin ^2 x}}$,we divide the numerator and the denominator by $x$ (since $x \rightarrow \infty$,$x > 0$,so $\sqrt{x^2} = x$):
$\lim _{x \rightarrow \infty} \frac{3 + \frac{4 \cos ^2 x}{x}}{\sqrt{\frac{x^2}{x^2} - \frac{5 \sin ^2 x}{x^2}}} = \lim _{x \rightarrow \infty} \frac{3 + \frac{4 \cos ^2 x}{x}}{\sqrt{1 - \frac{5 \sin ^2 x}{x^2}}}$
As $x \rightarrow \infty$,the terms $\frac{4 \cos ^2 x}{x} \rightarrow 0$ (because $\cos ^2 x$ is bounded between $0$ and $1$) and $\frac{5 \sin ^2 x}{x^2} \rightarrow 0$ (because $\sin ^2 x$ is bounded between $0$ and $1$).
Thus,the limit becomes $\frac{3 + 0}{\sqrt{1 - 0}} = \frac{3}{1} = 3$.

(A) We know that $\lim _{x \rightarrow 0} \frac{\sin(ax)}{ax} = 1$ and $1 - \cos(ax) = 2 \sin^2(\frac{ax}{2})$.
Given the expression: $\lim _{x \rightarrow 0} \frac{x^2 \sin ^2(3 x)+\sin ^4(6 x)}{(1-\cos 3 x)^2}$.
Using the identity $1 - \cos(3x) = 2 \sin^2(\frac{3x}{2})$,the denominator becomes $(2 \sin^2(\frac{3x}{2}))^2 = 4 \sin^4(\frac{3x}{2})$.
Now,divide numerator and denominator by $x^4$:
Numerator: $\frac{x^2 \sin^2(3x)}{x^4} + \frac{\sin^4(6x)}{x^4} = \frac{\sin^2(3x)}{x^2} + \frac{\sin^4(6x)}{x^4}$.
As $x \rightarrow 0$,$\frac{\sin^2(3x)}{x^2} \rightarrow (3)^2 = 9$ and $\frac{\sin^4(6x)}{x^4} \rightarrow (6)^4 = 1296$.
Denominator: $\frac{4 \sin^4(\frac{3x}{2})}{x^4} = 4 \cdot (\frac{\sin(\frac{3x}{2})}{\frac{3x}{2}})^4 \cdot (\frac{3}{2})^4 = 4 \cdot 1 \cdot \frac{81}{16} = \frac{81}{4}$.
Thus,the limit is $\frac{9 + 1296}{81/4} = \frac{1305 \times 4}{81} = \frac{145 \times 4}{9} = \frac{580}{9}$.

(D) We have $\operatorname{cosec} x - \cot x = \frac{1}{\sin x} - \frac{\cos x}{\sin x} = \frac{1-\cos x}{\sin x} = \tan(x/2)$.
Also,$e^x - e^{-x} = 2 \sinh x \approx 2x$ as $x \to 0$.
So,the numerator is $\tan(x/2) \cdot (e^x - e^{-x}) \approx (x/2) \cdot (2x) = x^2$.
For the denominator,$\sqrt{3} - \sqrt{2+\cos x} = \frac{3 - (2+\cos x)}{\sqrt{3} + \sqrt{2+\cos x}} = \frac{1-\cos x}{\sqrt{3} + \sqrt{2+\cos x}}$.
Using $1-\cos x \approx x^2/2$,the denominator is $\frac{x^2/2}{\sqrt{3} + \sqrt{3}} = \frac{x^2}{4\sqrt{3}}$.
Thus,the limit is $\lim_{x \to 0} \frac{x^2}{x^2 / (4\sqrt{3})} = 4\sqrt{3}$.

(B) We need to evaluate the limit: $L = \lim _{x \rightarrow 0} \frac{x \tan 2x - 2x \tan x}{(1 - \cos 2x)^2}$.
Using the identity $1 - \cos 2x = 2 \sin^2 x$,the denominator becomes $(2 \sin^2 x)^2 = 4 \sin^4 x$.
So,$L = \lim _{x \rightarrow 0} \frac{x \tan 2x - 2x \tan x}{4 \sin^4 x}$.
Using the Taylor series expansions for $\tan \theta \approx \theta + \frac{\theta^3}{3}$ and $\sin \theta \approx \theta$:
$\tan 2x \approx 2x + \frac{(2x)^3}{3} = 2x + \frac{8x^3}{3}$.
$\tan x \approx x + \frac{x^3}{3}$.
$\sin x \approx x$,so $\sin^4 x \approx x^4$.
Substituting these into the expression:
$L = \lim _{x \rightarrow 0} \frac{x(2x + \frac{8x^3}{3}) - 2x(x + \frac{x^3}{3})}{4x^4}$.
$L = \lim _{x \rightarrow 0} \frac{2x^2 + \frac{8x^4}{3} - 2x^2 - \frac{2x^4}{3}}{4x^4}$.
$L = \lim _{x \rightarrow 0} \frac{\frac{6x^4}{3}}{4x^4} = \lim _{x \rightarrow 0} \frac{2x^4}{4x^4} = \frac{2}{4} = \frac{1}{2}$.

(B) To evaluate the limit $\lim _{x \rightarrow \infty} \frac{(3-x)^{25}(6+x)^{35}}{(12+x)^{38}(9-x)^{22}}$,we consider the highest power of $x$ in the numerator and the denominator.
In the numerator,the term is $(3-x)^{25}(6+x)^{35} \approx (-x)^{25}(x)^{35} = -x^{60}$.
In the denominator,the term is $(12+x)^{38}(9-x)^{22} \approx (x)^{38}(-x)^{22} = x^{60}$.
Thus,the expression behaves as $\frac{-x^{60}}{x^{60}} = -1$ as $x \rightarrow \infty$.

(C) Let $L = \lim _{x \rightarrow \frac{\pi}{4}} \frac{2 \sqrt{2}-(\cos x+\sin x)^3}{1-\sin 2 x}$.
Substitute $x = \frac{\pi}{4} + h$,as $h \rightarrow 0$.
Then $\cos x + \sin x = \cos(\frac{\pi}{4} + h) + \sin(\frac{\pi}{4} + h) = \sqrt{2} \cos h$.
Also,$1 - \sin 2x = 1 - \sin(2(\frac{\pi}{4} + h)) = 1 - \sin(\frac{\pi}{2} + 2h) = 1 - \cos 2h = 2 \sin^2 h$.
Substituting these into the limit:
$L = \lim _{h \rightarrow 0} \frac{2 \sqrt{2} - (\sqrt{2} \cos h)^3}{2 \sin^2 h} = \lim _{h \rightarrow 0} \frac{2 \sqrt{2} - 2 \sqrt{2} \cos^3 h}{2 \sin^2 h} = \lim _{h \rightarrow 0} \frac{\sqrt{2}(1 - \cos^3 h)}{\sin^2 h}$.
Using $1 - \cos^3 h = (1 - \cos h)(1 + \cos h + \cos^2 h)$ and $\sin^2 h = (1 - \cos h)(1 + \cos h)$:
$L = \lim _{h \rightarrow 0} \frac{\sqrt{2}(1 - \cos h)(1 + \cos h + \cos^2 h)}{(1 - \cos h)(1 + \cos h)} = \lim _{h \rightarrow 0} \frac{\sqrt{2}(1 + \cos h + \cos^2 h)}{1 + \cos h}$.
As $h \rightarrow 0$,$\cos h \rightarrow 1$,so $L = \frac{\sqrt{2}(1 + 1 + 1)}{1 + 1} = \frac{3 \sqrt{2}}{2} = \frac{3}{\sqrt{2}}$.

(B) To evaluate the limit $\lim _{x \rightarrow 3} \frac{x^3-27}{x^2-9}$,we observe that substituting $x=3$ results in the indeterminate form $\frac{0}{0}$.
We factor the numerator using the difference of cubes formula $a^3-b^3 = (a-b)(a^2+ab+b^2)$ and the denominator using the difference of squares formula $a^2-b^2 = (a-b)(a+b)$.
$\lim _{x \rightarrow 3} \frac{(x-3)(x^2+3x+9)}{(x-3)(x+3)}$
Canceling the common factor $(x-3)$,we get:
$\lim _{x \rightarrow 3} \frac{x^2+3x+9}{x+3}$
Substituting $x=3$:
$\frac{3^2+3(3)+9}{3+3} = \frac{9+9+9}{6} = \frac{27}{6} = \frac{9}{2}$.

(A) Let $L = \lim _{x \rightarrow 1} \left( \frac{x+x^2+x^3+\ldots+x^n-n}{x-1} \right)$.
We can rewrite the numerator as: $(x-1) + (x^2-1) + (x^3-1) + \ldots + (x^n-1)$.
So,$L = \lim _{x \rightarrow 1} \left( \frac{(x-1) + (x^2-1) + (x^3-1) + \ldots + (x^n-1)}{x-1} \right)$.
$L = \lim _{x}$ ${\rightarrow 1} \left( \frac{x-1}{x-1} + \frac{x^2-1}{x-1} + \frac{x^3-1}{x-1} + \ldots + \frac{x^n-1}{x-1} \right)$.
Using the standard limit $\lim _{x \rightarrow a} \frac{x^n-a^n}{x-a} = na^{n-1}$,we get:
$L = 1 + 2(1)^{2-1} + 3(1)^{3-1} + \ldots + n(1)^{n-1}$.
$L = 1 + 2 + 3 + \ldots + n$.
$L = \frac{n(n+1)}{2}$.

(D) The limit $\lim _{x \rightarrow 0} \frac{1}{x}$ does not exist because the left-hand limit is $-\infty$ and the right-hand limit is $+\infty$.
Therefore,the assertion $(A)$ is false.
As $x$ approaches $0$ (decreases in magnitude),the value of $\frac{1}{x}$ increases in magnitude (tends to $\infty$ or $-\infty$). Thus,the reason $(R)$ is true.

(C) Given the limit expression: $\lim _{x \rightarrow 2} \frac{1+\sqrt{1+4 \log _2 x}}{2+\left(2 x+\sin ^2 x+2 \cos x\right)(2 x-4)}=m$
Since the denominator is non-zero at $x=2$ (specifically $2 + (4 + \sin^2 2 + 2 \cos 2)(0) = 2$),we can evaluate the limit by direct substitution:
$m = \frac{1+\sqrt{1+4 \log _2 2}}{2+\left(2(2)+\sin ^2 2+2 \cos 2\right)(2(2)-4)}$
$m = \frac{1+\sqrt{1+4(1)}}{2+(4+\sin ^2 2+2 \cos 2)(0)}$
$m = \frac{1+\sqrt{5}}{2}$
Now,calculate $m(m-1)$:
$m(m-1) = \left(\frac{1+\sqrt{5}}{2}\right) \left(\frac{1+\sqrt{5}}{2} - 1\right)$
$m(m-1) = \left(\frac{1+\sqrt{5}}{2}\right) \left(\frac{\sqrt{5}-1}{2}\right)$
Using the difference of squares formula $(a+b)(a-b) = a^2 - b^2$:
$m(m-1) = \frac{(\sqrt{5})^2 - (1)^2}{4} = \frac{5-1}{4} = \frac{4}{4} = 1$

(C) Let $f(x) = ax^2 + bx + c$. Since $l$ and $m$ are roots,$f(x) = a(x-l)(x-m)$.
If $\alpha \in (l, m)$,then $(x-l)$ and $(x-m)$ have opposite signs,so $f(x)$ has the opposite sign of $a$.
Thus,if $a > 0$,$f(x) < 0$,and if $a < 0$,$f(x) > 0$.
In both cases,$\frac{|f(x)|}{f(x)} = -1$ for $x \in (l, m)$.
Since $\frac{|a|}{a} = 1$ if $a > 0$ and $-1$ if $a < 0$,we have $\frac{-|a|}{a} = -1$ when $a > 0$ and $1$ when $a < 0$.
Wait,if $a > 0$,$\frac{-|a|}{a} = -1$. If $a < 0$,$\frac{-|a|}{a} = -(-a)/a = 1$. This matches the sign of $f(x)$ relative to $a$.
Therefore,$\lim_{x \rightarrow \alpha} \frac{|f(x)|}{f(x)} = \frac{-|a|}{a}$ when $\alpha \in (l, m)$.

(B) We know that $\cosh x = \frac{e^x + e^{-x}}{2}$.
Substituting this into the limit,we get $\lim _{x \rightarrow-\infty} \log _e \left( \frac{e^x + e^{-x}}{2} \right) + x$.
Using the property $\log(a/b) = \log a - \log b$,we have $\lim _{x \rightarrow-\infty} [\log _e(e^x + e^{-x}) - \log _e 2 + x]$.
Since $x \rightarrow -\infty$,$e^x \rightarrow 0$. We can factor out $e^{-x}$ inside the log: $\log _e(e^{-x}(e^{2x} + 1)) = \log _e(e^{-x}) + \log _e(1 + e^{2x}) = -x + \log _e(1 + e^{2x})$.
Substituting this back: $\lim _{x \rightarrow-\infty} [-x + \log _e(1 + e^{2x}) - \log _e 2 + x]$.
The terms $x$ and $-x$ cancel out,leaving $\lim _{x \rightarrow-\infty} [\log _e(1 + e^{2x}) - \log _e 2]$.
As $x \rightarrow -\infty$,$e^{2x} \rightarrow 0$,so $\log _e(1 + 0) - \log _e 2 = 0 - \log _e 2 = -\log _e 2$.

(A) Let $L = \lim_{x \rightarrow -\infty} \frac{3|x|-x}{|x|-2x} - \lim_{x \rightarrow 0} \frac{\log(1+x^3)}{\sin^3 x}$.
For the first limit,as $x \rightarrow -\infty$,$|x| = -x$. Thus,$\lim_{x \rightarrow -\infty} \frac{3(-x)-x}{-x-2x} = \lim_{x \rightarrow -\infty} \frac{-4x}{-3x} = \frac{4}{3}$.
For the second limit,we use the standard limits $\lim_{x \rightarrow 0} \frac{\log(1+x^3)}{x^3} = 1$ and $\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$.
So,$\lim_{x \rightarrow 0} \frac{\log(1+x^3)}{\sin^3 x} = \lim_{x \rightarrow 0} \left( \frac{\log(1+x^3)}{x^3} \times \frac{x^3}{\sin^3 x} \right) = 1 \times 1 = 1$.
Therefore,$L = \frac{4}{3} - 1 = \frac{1}{3}$.

(D) Let $f(x) = \frac{\sqrt{11+|x|-6 \sqrt{2+|x|}}}{6-2 \sqrt{2+|x|}}$.
We observe that $11+|x|-6 \sqrt{2+|x|} = 9 + 2 + |x| - 6\sqrt{2+|x|} = 3^2 + (\sqrt{2+|x|})^2 - 2(3)(\sqrt{2+|x|}) = (3-\sqrt{2+|x|})^2$.
Substituting this into the limit expression:
$\lim _{x \rightarrow 0} \frac{\sqrt{(3-\sqrt{2+|x|})^2}}{2(3-\sqrt{2+|x|})}$
$= \lim _{x \rightarrow 0} \frac{|3-\sqrt{2+|x|}|}{2(3-\sqrt{2+|x|})}$.
As $x \rightarrow 0$,$\sqrt{2+|x|} \rightarrow \sqrt{2} \approx 1.414$,so $3-\sqrt{2+|x|} > 0$.
Thus,$|3-\sqrt{2+|x|}| = 3-\sqrt{2+|x|}$.
$= \lim _{x \rightarrow 0} \frac{3-\sqrt{2+|x|}}{2(3-\sqrt{2+|x|})} = \frac{1}{2}$.

(A) Let $L = \lim _{x \rightarrow 0} x^3 \left( \sqrt{x^2 + \sqrt{x^4 + 1}} - \sqrt{2} x \right)$.
As $x \rightarrow 0$,the expression inside the limit becomes $0 \times (\sqrt{0 + \sqrt{0 + 1}} - 0) = 0 \times (1 - 0) = 0$.
Since the expression is $0 \times 1$,the limit evaluates directly to $0$.
Thus,$\lim _{x \rightarrow 0} x^3 \left( \sqrt{x^2 + \sqrt{x^4 + 1}} - \sqrt{2} x \right) = 0 \times (1 - 0) = 0$.

(A) We are given the limit $L = \lim _{x \rightarrow 0} \frac{x^2 \log (\cos x)}{\log (1+x^2)}$.
This is a $\frac{0}{0}$ indeterminate form.
We use the standard limits $\lim _{u \rightarrow 0} \frac{\log (1+u)}{u} = 1$ and the Taylor series expansion for $\log(\cos x)$.
First,$\log(1+x^2) \approx x^2$ as $x \rightarrow 0$.
Next,$\cos x \approx 1 - \frac{x^2}{2}$,so $\log(\cos x) \approx \log(1 - \frac{x^2}{2}) \approx -\frac{x^2}{2}$.
Substituting these into the limit:
$L = \lim _{x \rightarrow 0} \frac{x^2 \cdot (-\frac{x^2}{2})}{x^2} = \lim _{x \rightarrow 0} (-\frac{x^2}{2}) = 0$.

(D) We are given the limit $\lim _{x \rightarrow \frac{\pi^{+}}{2}} \frac{[\sin x]-[\cos x]+1}{2}$.
As $x \rightarrow \frac{\pi^{+}}{2}$,$x$ is slightly greater than $\frac{\pi}{2}$.
For $x$ in the interval $(\frac{\pi}{2}, \pi)$,we have $0 \leq \sin x < 1$,so $[\sin x] = 0$.
Also,for $x$ in the interval $(\frac{\pi}{2}, \pi)$,we have $-1 < \cos x < 0$,so $[\cos x] = -1$.
Substituting these values into the expression:
$\lim _{x \rightarrow \frac{\pi^{+}}{2}} \frac{0 - (-1) + 1}{2} = \frac{1 + 1}{2} = \frac{2}{2} = 1$.

(D) To evaluate the limit $\lim _{x \rightarrow \infty} \frac{11 x^3-3 x+4}{13 x^3-5 x^2-7}$,we divide the numerator and the denominator by the highest power of $x$,which is $x^3$:
$\lim _{x \rightarrow \infty} \frac{x^3(11 - \frac{3}{x^2} + \frac{4}{x^3})}{x^3(13 - \frac{5}{x} - \frac{7}{x^3})}$
$\lim _{x \rightarrow \infty} \frac{11 - \frac{3}{x^2} + \frac{4}{x^3}}{13 - \frac{5}{x} - \frac{7}{x^3}}$
As $x \rightarrow \infty$,the terms $\frac{3}{x^2}, \frac{4}{x^3}, \frac{5}{x},$ and $\frac{7}{x^3}$ all approach $0$.
Therefore,the limit is $\frac{11 - 0 + 0}{13 - 0 - 0} = \frac{11}{13}$.
Comparing this to $\frac{a}{b}$,we get $a = 11$ and $b = 13$.
Thus,$a + b = 11 + 13 = 24$.

(B) Let $L = \lim _{x \rightarrow 1} \frac{(1-x)(1-x^2) \cdots (1-x^{2n})}{\{(1-x)(1-x^2) \cdots (1-x^n)\}^2}$.
We can rewrite the expression as:
$L = \lim _{x \rightarrow 1} \frac{\prod_{k=1}^{2n} (1-x^k)}{\left(\prod_{k=1}^{n} (1-x^k)\right)^2}$.
Using the property $\lim _{x \rightarrow 1} \frac{1-x^k}{1-x} = k$,we divide the numerator and denominator by $(1-x)^{2n}$:
$L = \lim _{x \rightarrow 1} \frac{\prod_{k=1}^{2n} \frac{1-x^k}{1-x}}{\left(\prod_{k=1}^{n} \frac{1-x^k}{1-x}\right)^2}$.
As $x \rightarrow 1$,$\frac{1-x^k}{1-x} = 1 + x + x^2 + \dots + x^{k-1} \rightarrow k$.
Substituting these limits:
$L = \frac{1 \times 2 \times 3 \times \dots \times 2n}{(1 \times 2 \times 3 \times \dots \times n)^2}$.
$L = \frac{(2n)!}{(n!)^2} = \frac{(2n)!}{n! n!} = {}^{2n}C_n$.

(A) We are given the limit $\lim _{x \rightarrow 0} \frac{a^x-1}{\sin x}$.
Dividing both the numerator and the denominator by $x$,we get:
$\lim _{x \rightarrow 0} \frac{\frac{a^x-1}{x}}{\frac{\sin x}{x}}$.
Using the standard limits $\lim _{x \rightarrow 0} \frac{a^x-1}{x} = \log _e a$ and $\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1$,we have:
$\frac{\log _e a}{1} = \log _e a$.
Thus,the correct option is $A$.

(B) To evaluate the limit $\lim _{n \rightarrow \infty}\left\{n-\sqrt{n^2-4 n}\right\}$,we multiply and divide by the conjugate expression:
$= \lim _{n \rightarrow \infty} \frac{(n-\sqrt{n^2-4 n})(n+\sqrt{n^2-4 n})}{n+\sqrt{n^2-4 n}}$
$= \lim _{n \rightarrow \infty} \frac{n^2-(n^2-4 n)}{n+\sqrt{n^2-4 n}}$
$= \lim _{n \rightarrow \infty} \frac{4 n}{n+\sqrt{n^2-4 n}}$
Divide the numerator and denominator by $n$:
$= \lim _{n \rightarrow \infty} \frac{4}{1+\sqrt{1-\frac{4}{n}}}$
As $n \rightarrow \infty$,$\frac{4}{n} \rightarrow 0$,so the expression becomes:
$= \frac{4}{1+\sqrt{1-0}} = \frac{4}{1+1} = \frac{4}{2} = 2$.

(C) To find $\alpha$:
$\alpha = \lim _{x \rightarrow 0} \frac{x(2^x-1)}{1-\cos x} = \lim _{x \rightarrow 0} \frac{x(2^x-1)}{2 \sin^2(x/2)}$
$= \lim _{x \rightarrow 0} \frac{1}{2} \cdot \frac{2^x-1}{x} \cdot \frac{x^2}{\sin^2(x/2)} \cdot \frac{1}{x} = \lim _{x \rightarrow 0} \frac{1}{2} \cdot \frac{2^x-1}{x} \cdot \frac{1}{(\frac{\sin(x/2)}{x/2})^2 \cdot \frac{1}{4}} = \frac{1}{2} \cdot \ln 2 \cdot 4 = 2 \ln 2$
So,$\alpha = 2 \ln 2$ ... $(i)$
To find $\beta$:
$\beta = \lim _{x \rightarrow 0} \frac{x(2^x-1)}{\sqrt{1+x^2}-\sqrt{1-x^2}}$
Rationalizing the denominator:
$\beta = \lim _{x \rightarrow 0} \frac{x(2^x-1)(\sqrt{1+x^2}+\sqrt{1-x^2})}{(1+x^2)-(1-x^2)} = \lim _{x \rightarrow 0} \frac{x(2^x-1)(\sqrt{1+x^2}+\sqrt{1-x^2})}{2x^2}$
$= \lim _{x \rightarrow 0} \frac{1}{2} \cdot \frac{2^x-1}{x} \cdot (\sqrt{1+x^2}+\sqrt{1-x^2}) = \frac{1}{2} \cdot \ln 2 \cdot (1+1) = \ln 2$
So,$\beta = \ln 2$ ... $(ii)$
From $(i)$ and $(ii)$,we get $\alpha = 2\beta$.

(D) We evaluate the limit as $x \rightarrow \infty$:
$\lim _{x \rightarrow \infty}\left(\frac{6 x^2-\cos 3 x}{x^2+5}-\frac{5 x^3+3}{\sqrt{x^6+2}}\right)$
Divide the numerator and denominator of the first term by $x^2$:
$\lim _{x \rightarrow \infty} \frac{6 - \frac{\cos 3x}{x^2}}{1 + \frac{5}{x^2}} = \frac{6 - 0}{1 + 0} = 6$
For the second term,since $x \rightarrow \infty$,$x > 0$,so $\sqrt{x^6} = x^3$:
$\lim _{x \rightarrow \infty} \frac{5x^3 + 3}{\sqrt{x^6+2}} = \lim _{x \rightarrow \infty} \frac{5 + \frac{3}{x^3}}{\sqrt{1 + \frac{2}{x^6}}} = \frac{5 + 0}{\sqrt{1 + 0}} = 5$
Thus,the limit is $6 - 5 = 1$.

(C) To evaluate the limit $\lim _{x \rightarrow \infty}\left(\sqrt{x^2+a x+b}-x\right)$,we rationalize the expression:
$= \lim _{x}$ ${\rightarrow \infty}\left(\left(\sqrt{x^2+a x+b}-x\right) \times \frac{\sqrt{x^2+a x+b}+x}{\sqrt{x^2+a x+b}+x}\right)$
$= \lim _{x \rightarrow \infty}\left(\frac{x^2+a x+b-x^2}{\sqrt{x^2+a x+b}+x}\right)$
$= \lim _{x \rightarrow \infty}\left(\frac{a x+b}{\sqrt{x^2+a x+b}+x}\right)$
Divide the numerator and denominator by $x$:
$= \lim _{x \rightarrow \infty} \frac{a+\frac{b}{x}}{\sqrt{1+\frac{a}{x}+\frac{b}{x^2}}+1}$
As $x \rightarrow \infty$,$\frac{1}{x} \rightarrow 0$ and $\frac{1}{x^2} \rightarrow 0$,so the limit becomes:
$= \frac{a+0}{\sqrt{1+0+0}+1} = \frac{a}{2}$
Since the result is $\frac{a}{2}$,the limit depends only on $a$.

(B) Let the expression be $L = \lim _{n \rightarrow \infty} \frac{\left[6^2(1^2+2^2+\ldots+n^2)\right]^2}{[5(1+2+\ldots+n)]\left[2^3(1^3+2^3+\ldots+n^3)\right]}$.
Using the summation formulas $\sum_{k=1}^n k = \frac{n(n+1)}{2}$,$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$,and $\sum_{k=1}^n k^3 = \frac{n^2(n+1)^2}{4}$:
Numerator $= 36^2 \left[\frac{n(n+1)(2n+1)}{6}\right]^2 = 1296 \frac{n^2(n+1)^2(2n+1)^2}{36} = 36 n^2(n+1)^2(2n+1)^2$.
Denominator $= 5 \left[\frac{n(n+1)}{2}\right] \times 8 \left[\frac{n^2(n+1)^2}{4}\right] = 5 \times \frac{n(n+1)}{2} \times 2n^2(n+1)^2 = 5 n^3(n+1)^3$.
$L = \lim _{n \rightarrow \infty} \frac{36 n^2(n+1)^2(2n+1)^2}{5 n^3(n+1)^3} = \lim _{n \rightarrow \infty} \frac{36 n^2(n+1)^2(4n^2)}{5 n^3(n+1)^3} = \lim _{n \rightarrow \infty} \frac{144 n^6}{5 n^6} = \frac{144}{5}$.

(B) We need to evaluate the limit: $L = \lim_{x \rightarrow 0} \left( \frac{4^x - 1}{2^x - 1} - \frac{\sqrt{4 + 3x} - 2}{x} \right)$.
First,consider the first term: $\lim_{x \rightarrow 0} \frac{4^x - 1}{2^x - 1} = \lim_{x \rightarrow 0} \frac{(2^x - 1)(2^x + 1)}{2^x - 1} = \lim_{x \rightarrow 0} (2^x + 1) = 2^0 + 1 = 1 + 1 = 2$.
Next,consider the second term: $\lim_{x \rightarrow 0} \frac{\sqrt{4 + 3x} - 2}{x}$.
Rationalizing the numerator: $\lim_{x \rightarrow 0} \frac{(\sqrt{4 + 3x} - 2)(\sqrt{4 + 3x} + 2)}{x(\sqrt{4 + 3x} + 2)} = \lim_{x \rightarrow 0} \frac{4 + 3x - 4}{x(\sqrt{4 + 3x} + 2)} = \lim_{x \rightarrow 0} \frac{3x}{x(\sqrt{4 + 3x} + 2)} = \lim_{x \rightarrow 0} \frac{3}{\sqrt{4 + 3x} + 2} = \frac{3}{\sqrt{4} + 2} = \frac{3}{2 + 2} = \frac{3}{4}$.
Subtracting the two results: $2 - \frac{3}{4} = \frac{8 - 3}{4} = \frac{5}{4}$.

(C) First,evaluate $l$:
$\lim _{x \rightarrow 0} \frac{e^{2 x}-1}{5 x} = \frac{1}{5} \lim _{x \rightarrow 0} \frac{e^{2 x}-1}{x} = \frac{1}{5} \times 2 = \frac{2}{5}$.
So,$l = \frac{2}{5}$.
Then,$5l = 5 \times \frac{2}{5} = 2$.
Next,evaluate $m$:
$\lim _{x \rightarrow 1} \frac{2 \log x}{x-1}$.
Let $x = 1+h$,as $x \rightarrow 1, h \rightarrow 0$.
$\lim _{h \rightarrow 0} \frac{2 \log(1+h)}{h} = 2(1) = 2$.
So,$m = 2$.
The roots of the cubic equation are $2, 2$,and $1$.
The equation is $(x-2)(x-2)(x-1) = 0$.
$(x^2-4x+4)(x-1) = 0$.
$x^3 - x^2 - 4x^2 + 4x + 4x - 4 = 0$.
$x^3 - 5x^2 + 8x - 4 = 0$.

(C) Given,$g(x) = \frac{x}{[x]}$.
For $x > 2$ and $x$ close to $2$,$[x] = 2$,so $g(x) = \frac{x}{2}$.
Also,$g(2) = \frac{2}{[2]} = \frac{2}{2} = 1$.
Now,we evaluate the right-hand limit:
$\lim_{x \rightarrow 2^+} \frac{g(x) - g(2)}{x - 2} = \lim_{x \rightarrow 2^+} \frac{\frac{x}{2} - 1}{x - 2}$
$= \lim_{x \rightarrow 2^+} \frac{x - 2}{2(x - 2)}$
$= \lim_{x \rightarrow 2^+} \frac{1}{2} = \frac{1}{2}$.

(B) We know that $\sin(3\theta) = 3 \sin \theta - 4 \sin^3 \theta$.
Thus,$l = \lim_{\theta \rightarrow 0} \frac{\sin(3\theta)}{\theta} = \lim_{\theta \rightarrow 0} \frac{\sin(3\theta)}{3\theta} \times 3 = 1 \times 3 = 3$.
We also know that $\tan(2\theta) = \frac{2 \tan \theta}{1 - \tan^2 \theta}$.
Thus,$m = \lim_{\theta \rightarrow 0} \frac{\tan(2\theta)}{\theta} = \lim_{\theta \rightarrow 0} \frac{\tan(2\theta)}{2\theta} \times 2 = 1 \times 2 = 2$.
The quadratic equation with roots $l=3$ and $m=2$ is given by $(x - l)(x - m) = 0$.
$(x - 3)(x - 2) = 0$
$x^2 - 2x - 3x + 6 = 0$
$x^2 - 5x + 6 = 0$.

(B) Given the limit $\lim _{x \rightarrow 2^{+}}\left(\frac{[x]^3}{3}-\left[\frac{x}{3}\right]^3\right)$.
As $x \rightarrow 2^{+}$,$x$ is slightly greater than $2$,so $[x] = 2$.
Also,as $x \rightarrow 2^{+}$,$\frac{x}{3}$ is slightly greater than $\frac{2}{3}$,so $\left[\frac{x}{3}\right] = 0$.
Substituting these values into the expression:
$\lim _{x}$ ${\rightarrow 2^{+}}\left(\frac{[x]^3}{3}-\left[\frac{x}{3}\right]^3\right) = \frac{2^3}{3} - 0^3 = \frac{8}{3} - 0 = \frac{8}{3}$.

(B) We need to evaluate the limit $L = \lim _{x \rightarrow 0} \frac{1-\cos x \cos 2 x}{\sin ^2 x}$.
Using the identity $\cos 2x = 1 - 2\sin^2 x$,we rewrite the numerator:
$1 - \cos x(1 - 2\sin^2 x) = 1 - \cos x + 2\sin^2 x \cos x$.
So,$L = \lim _{x \rightarrow 0} \frac{1 - \cos x + 2\sin^2 x \cos x}{\sin^2 x}$.
$L = \lim _{x \rightarrow 0} \left( \frac{1 - \cos x}{\sin^2 x} + \frac{2\sin^2 x \cos x}{\sin^2 x} \right)$.
Using $\sin^2 x = 1 - \cos^2 x = (1 - \cos x)(1 + \cos x)$:
$L = \lim _{x \rightarrow 0} \left( \frac{1 - \cos x}{(1 - \cos x)(1 + \cos x)} + 2\cos x \right)$.
$L = \lim _{x \rightarrow 0} \left( \frac{1}{1 + \cos x} + 2\cos x \right)$.
Substituting $x = 0$:
$L = \frac{1}{1 + 1} + 2(1) = \frac{1}{2} + 2 = \frac{5}{2}$.

(C) Given $f(x) = \frac{5x \operatorname{cosec}(\sqrt{x}) - 1}{(x - 2) \operatorname{cosec}(\sqrt{x})}$.
Substituting $x^2$ for $x$,we get:
$f(x^2) = \frac{5x^2 \operatorname{cosec}(x) - 1}{(x^2 - 2) \operatorname{cosec}(x)} = \frac{5x^2 \operatorname{cosec}(x)}{(x^2 - 2) \operatorname{cosec}(x)} - \frac{1}{(x^2 - 2) \operatorname{cosec}(x)}$.
$f(x^2) = \frac{5x^2}{x^2 - 2} - \frac{\sin(x)}{x^2 - 2}$.
Now,taking the limit as $x \rightarrow \infty$:
$\lim_{x \rightarrow \infty} f(x^2) = \lim_{x \rightarrow \infty} \left( \frac{5x^2}{x^2 - 2} - \frac{\sin(x)}{x^2 - 2} \right)$.
Since $\lim_{x \rightarrow \infty} \frac{5x^2}{x^2 - 2} = 5$ and $\lim_{x \rightarrow \infty} \frac{\sin(x)}{x^2 - 2} = 0$ (as $-1 \leq \sin(x) \leq 1$ and $x^2 - 2 \rightarrow \infty$),
$\lim_{x \rightarrow \infty} f(x^2) = 5 - 0 = 5$.

(A) Let $L = \lim _{x \rightarrow \frac{\pi}{2}} \tan(\frac{\pi}{4} - \frac{x}{2}) \cdot \frac{1-\sin x}{(\pi-2 x)^3}$.
Substitute $x = \frac{\pi}{2} + h$,where $h \rightarrow 0$ as $x \rightarrow \frac{\pi}{2}$.
Then $\frac{\pi}{4} - \frac{x}{2} = \frac{\pi}{4} - \frac{\pi}{4} - \frac{h}{2} = -\frac{h}{2}$.
Also,$\pi - 2x = \pi - 2(\frac{\pi}{2} + h) = -2h$.
$1 - \sin x = 1 - \sin(\frac{\pi}{2} + h) = 1 - \cos h = 2\sin^2(\frac{h}{2})$.
Substituting these into the limit:
$L = \lim _{h \rightarrow 0} \tan(-\frac{h}{2}) \cdot \frac{2\sin^2(\frac{h}{2})}{(-2h)^3}$.
$L = \lim _{h \rightarrow 0} -\tan(\frac{h}{2}) \cdot \frac{2\sin^2(\frac{h}{2})}{-8h^3}$.
$L = \lim _{h \rightarrow 0} \frac{\tan(\frac{h}{2})}{h} \cdot \frac{2\sin^2(\frac{h}{2})}{8h^2}$.
$L = \lim _{h}$ ${\rightarrow 0} \frac{1}{2} \cdot \frac{\tan(\frac{h}{2})}{\frac{h}{2}} \cdot \frac{2}{8} \cdot (\frac{\sin(\frac{h}{2})}{\frac{h}{2}})^2 \cdot \frac{1}{4}$.
$L = \frac{1}{2} \cdot 1 \cdot \frac{1}{4} \cdot 1^2 \cdot \frac{1}{4} = \frac{1}{32}$.

(D) We are given the limit: $\lim _{n \rightarrow \infty} \frac{\cos x - e^{nx}}{1 - A e^{nx}}$.
Divide the numerator and the denominator by $e^{nx}$:
$= \lim _{n \rightarrow \infty} \frac{\frac{\cos x}{e^{nx}} - 1}{\frac{1}{e^{nx}} - A}$.
Since $x > 0$,as $n \rightarrow \infty$,$e^{nx} \rightarrow \infty$.
Thus,$\frac{\cos x}{e^{nx}} \rightarrow 0$ and $\frac{1}{e^{nx}} \rightarrow 0$.
Substituting these values,we get:
$= \frac{0 - 1}{0 - A} = \frac{-1}{-A} = \frac{1}{A}$.

(C) Given limit is $\lim _{n \rightarrow \infty} \frac{n(2 n+1)^2}{(n+2)(n^2+3 n-1)}$.
Divide the numerator and denominator by the highest power of $n$,which is $n^3$.
Numerator: $n(2n+1)^2 = n(4n^2+4n+1) = 4n^3+4n^2+n$.
Denominator: $(n+2)(n^2+3n-1) = n^3+3n^2-n+2n^2+6n-2 = n^3+5n^2+5n-2$.
Now,$\lim _{n \rightarrow \infty} \frac{4n^3+4n^2+n}{n^3+5n^2+5n-2} = \lim _{n \rightarrow \infty} \frac{4 + \frac{4}{n} + \frac{1}{n^2}}{1 + \frac{5}{n} + \frac{5}{n^2} - \frac{2}{n^3}}$.
As $n \rightarrow \infty$,$\frac{1}{n}, \frac{1}{n^2}, \frac{1}{n^3} \rightarrow 0$.
Therefore,the limit is $\frac{4+0+0}{1+0+0-0} = 4$.

(C) Let $L = \lim _{x \rightarrow 1} \frac{\sqrt{1-\cos 2(x-1)}}{x-1}$.
Using the identity $1-\cos 2\theta = 2\sin^2\theta$,we have:
$L = \lim _{x \rightarrow 1} \frac{\sqrt{2\sin^2(x-1)}}{x-1} = \sqrt{2} \lim _{x \rightarrow 1} \frac{|\sin(x-1)|}{x-1}$.
Let $z = x-1$. As $x \rightarrow 1$,$z \rightarrow 0$.
$L = \sqrt{2} \lim _{z \rightarrow 0} \frac{|\sin z|}{z}$.
Now,evaluate the one-sided limits:
Right-hand limit $(RHL)$: $\sqrt{2} \lim _{z \rightarrow 0^+} \frac{\sin z}{z} = \sqrt{2}(1) = \sqrt{2}$.
Left-hand limit $(LHL)$: $\sqrt{2} \lim _{z \rightarrow 0^-} \frac{-\sin z}{z} = \sqrt{2}(-1) = -\sqrt{2}$.
Since $RHL \neq LHL$,the limit does not exist.

(D) Let $L = \lim _{x}$ ${\rightarrow 0} \frac{8}{x^8}\left[1-\cos \left(\frac{x^2}{2}\right)-\cos \left(\frac{x^2}{4}\right)+\cos \left(\frac{x^2}{2}\right) \cdot \cos \left(\frac{x^2}{4}\right)\right]$
Factor the expression inside the bracket:
$L = \lim _{x}$ ${\rightarrow 0} \frac{8}{x^8}\left[\left(1-\cos \frac{x^2}{2}\right) - \cos \left(\frac{x^2}{4}\right)\left(1-\cos \frac{x^2}{2}\right)\right]$
$L = \lim _{x \rightarrow 0} \frac{8}{x^8}\left(1-\cos \frac{x^2}{2}\right)\left(1-\cos \frac{x^2}{4}\right)$
Using the identity $1-\cos \theta = 2 \sin^2 \left(\frac{\theta}{2}\right)$:
$L = \lim _{x \rightarrow 0} \frac{8}{x^8} \left(2 \sin^2 \frac{x^2}{4}\right) \left(2 \sin^2 \frac{x^2}{8}\right)$
$L = \lim _{x \rightarrow 0} \frac{32}{x^8} \left(\sin^2 \frac{x^2}{4}\right) \left(\sin^2 \frac{x^2}{8}\right)$
Multiply and divide by $\left(\frac{x^2}{4}\right)^2$ and $\left(\frac{x^2}{8}\right)^2$:
$L = 32 \lim _{x}$ ${\rightarrow 0} \left(\frac{\sin \frac{x^2}{4}}{\frac{x^2}{4}}\right)^2 \left(\frac{x^4}{16}\right) \left(\frac{\sin \frac{x^2}{8}}{\frac{x^2}{8}}\right)^2 \left(\frac{x^4}{64}\right) \cdot \frac{1}{x^8}$
$L = 32 \cdot (1)^2 \cdot \frac{1}{16} \cdot (1)^2 \cdot \frac{1}{64}$
$L = 32 \cdot \frac{1}{1024} = \frac{1}{32}$

(D) Given the limit: $\lim _{x \rightarrow -3} \frac{\sin ^{-1}(x+3)}{x^2+3x}$
Substitute $t = x+3$. As $x \rightarrow -3$,$t \rightarrow 0$.
Then $x = t-3$.
The expression becomes: $\lim _{t \rightarrow 0} \frac{\sin ^{-1}(t)}{(t-3)t} = \lim _{t \rightarrow 0} \left( \frac{\sin ^{-1}(t)}{t} \right) \cdot \frac{1}{t-3}$
Using the standard limit $\lim _{t \rightarrow 0} \frac{\sin ^{-1}(t)}{t} = 1$:
$= 1 \cdot \frac{1}{0-3} = -\frac{1}{3}$

(B) We know that $[r^2 x] = r^2 x - \{r^2 x\}$,where $\{r^2 x\}$ is the fractional part of $r^2 x$.
Substituting this into the limit,we get:
$\lim_{n \rightarrow \infty} \frac{1}{n^3} \sum_{r=1}^n [r^2 x] = \lim_{n \rightarrow \infty} \frac{1}{n^3} \sum_{r=1}^n (r^2 x - \{r^2 x\})$
$= \lim_{n \rightarrow \infty} \left( \frac{x}{n^3} \sum_{r=1}^n r^2 - \frac{1}{n^3} \sum_{r=1}^n \{r^2 x\} \right)$
Using the formula $\sum_{r=1}^n r^2 = \frac{n(n+1)(2n+1)}{6}$,we have:
$= \lim_{n \rightarrow \infty} \left( \frac{x \cdot n(n+1)(2n+1)}{6n^3} - \frac{1}{n^3} \sum_{r=1}^n \{r^2 x\} \right)$
Since $0 \leq \{r^2 x\} < 1$,the second term $\frac{1}{n^3} \sum_{r=1}^n \{r^2 x\}$ is bounded by $\frac{n}{n^3} = \frac{1}{n^2}$,which approaches $0$ as $n \rightarrow \infty$.
Thus,the limit is $\lim_{n \rightarrow \infty} \frac{x(2n^3 + 3n^2 + n)}{6n^3} = \frac{2x}{6} = \frac{x}{3}$.

(B) Let $L = \cos \left[ \lim_{x \rightarrow \infty} \frac{2 \pi |x| + \pi x}{|x| - 3x} + \lim_{x \rightarrow 0} \frac{\cos \left( \frac{\pi}{2} \cos^2 x \right)}{x^2} \right]$.
For the first limit,as $x \rightarrow \infty$,$|x| = x$. Thus,$\lim_{x \rightarrow \infty} \frac{2 \pi x + \pi x}{x - 3x} = \lim_{x \rightarrow \infty} \frac{3 \pi x}{-2x} = -\frac{3 \pi}{2}$.
For the second limit,$\lim_{x \rightarrow 0} \frac{\cos \left( \frac{\pi}{2} \cos^2 x \right)}{x^2} = \lim_{x \rightarrow 0} \frac{\cos \left( \frac{\pi}{2} (1 - \sin^2 x) \right)}{x^2} = \lim_{x \rightarrow 0} \frac{\cos \left( \frac{\pi}{2} - \frac{\pi}{2} \sin^2 x \right)}{x^2}$.
Using $\cos \left( \frac{\pi}{2} - \theta \right) = \sin \theta$,we get $\lim_{x \rightarrow 0} \frac{\sin \left( \frac{\pi}{2} \sin^2 x \right)}{x^2}$.
Since $\sin \theta \approx \theta$ for small $\theta$,this becomes $\lim_{x \rightarrow 0} \frac{\frac{\pi}{2} \sin^2 x}{x^2} = \frac{\pi}{2} \times (1)^2 = \frac{\pi}{2}$.
Substituting these values back into the expression: $L = \cos \left( -\frac{3 \pi}{2} + \frac{\pi}{2} \right) = \cos(-\pi) = -1$.

(D) The given limit is $\lim _{x \rightarrow 0} \frac{\sqrt{1+x \sin x}-\sqrt{\cos x}}{\tan ^2 2 x}$,which is of the $\frac{0}{0}$ form.
Rationalizing the numerator:
$= \lim _{x}$ ${\rightarrow 0} \frac{(\sqrt{1+x \sin x}-\sqrt{\cos x})(\sqrt{1+x \sin x}+\sqrt{\cos x})}{\tan ^2 2 x (\sqrt{1+x \sin x}+\sqrt{\cos x})}$
$= \lim _{x \rightarrow 0} \frac{1+x \sin x-\cos x}{\tan ^2 2 x} \cdot \frac{1}{\sqrt{1+x \sin x}+\sqrt{\cos x}} $
$= \lim _{x \rightarrow 0} \frac{(1-\cos x)+x \sin x}{\tan ^2 2 x} \cdot \frac{1}{2} $
Using the identities $1-\cos x = 2 \sin ^2 \frac{x}{2}$ and $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$:
$= \lim _{x}$ ${\rightarrow 0} \frac{2 \sin ^2 \frac{x}{2} + 2x \sin \frac{x}{2} \cos \frac{x}{2}}{\tan ^2 2 x} \cdot \frac{1}{2}$
$= \lim _{x}$ ${\rightarrow 0} \frac{2 \sin ^2 \frac{x}{2} [1 + \frac{x \cos \frac{x}{2}}{\sin \frac{x}{2}}]}{\tan ^2 2 x} \cdot \frac{1}{2}$
$= \lim _{x \rightarrow 0} \frac{2 \sin ^2 \frac{x}{2} [1 + \frac{x}{\tan \frac{x}{2}}]}{\tan ^2 2 x} \cdot \frac{1}{2} $
$= \lim _{x \rightarrow 0} \frac{2 (\frac{x}{2})^2 [1 + 2 \cdot \frac{x/2}{\tan (x/2)}]}{(2x)^2} \cdot \frac{1}{2} $
$= \lim _{x}$ ${\rightarrow 0} \frac{2 \cdot \frac{x^2}{4} [1+2]}{4x^2} \cdot \frac{1}{2} = \frac{1/2 \cdot 3}{4} \cdot \frac{1}{2} = \frac{3}{16}$.

(B) Given $\Delta(x) = \begin{vmatrix} e^x & -1 \\ \sin x - 1 & 1 \end{vmatrix}$.
Expanding the determinant,we get $\Delta(x) = (e^x)(1) - (-1)(\sin x - 1)$.
$\Delta(x) = e^x + \sin x - 1$.
Now,we need to evaluate $\lim_{x \rightarrow 0} \frac{\Delta(x)}{x} = \lim_{x \rightarrow 0} \frac{e^x + \sin x - 1}{x}$.
This is a $\frac{0}{0}$ form,so we use standard limits.
Using standard limits: $\lim_{x \rightarrow 0} \left( \frac{e^x - 1}{x} + \frac{\sin x}{x} \right)$.
$= 1 + 1 = 2$.

(D) We know that $\cosh x = \frac{e^x + e^{-x}}{2}$.
Substituting this into the limit,we get:
$\lim _{x \rightarrow \infty} [x - \log (\frac{e^x + e^{-x}}{2})]$
$= \lim _{x \rightarrow \infty} [x - (\log (e^x + e^{-x}) - \log 2)]$
$= \lim _{x \rightarrow \infty} [x - \log (e^x(1 + e^{-2x})) + \log 2]$
$= \lim _{x \rightarrow \infty} [x - (\log e^x + \log (1 + e^{-2x})) + \log 2]$
$= \lim _{x \rightarrow \infty} [x - x - \log (1 + e^{-2x}) + \log 2]$
$= \lim _{x \rightarrow \infty} [\log 2 - \log (1 + e^{-2x})]$
As $x \rightarrow \infty$,$e^{-2x} \rightarrow 0$,so $\log (1 + e^{-2x}) \rightarrow \log 1 = 0$.
Therefore,the limit is $\log 2 - 0 = \log 2$.