Concept of limits, Evaluation of algebric limits Questions in English

(B) We need to evaluate $\lim_{x \rightarrow \pi/4} \frac{\cot^3 x - \tan x}{\cos(x + \pi/4)}$.
Let $t = \tan x$. As $x \rightarrow \pi/4$,$t \rightarrow 1$.
The numerator is $\frac{1}{t^3} - t = \frac{1 - t^4}{t^3} = \frac{(1 - t^2)(1 + t^2)}{t^3} = \frac{(1 - t)(1 + t)(1 + t^2)}{t^3}$.
The denominator is $\cos(x + \pi/4) = \cos x \cos(\pi/4) - \sin x \sin(\pi/4) = \frac{1}{\sqrt{2}}(\cos x - \sin x) = \frac{\cos x}{\sqrt{2}}(1 - \tan x) = \frac{\cos x}{\sqrt{2}}(1 - t)$.
Substituting these into the limit:
$\lim_{t \rightarrow 1} \frac{(1 - t)(1 + t)(1 + t^2)}{t^3} \cdot \frac{\sqrt{2}}{\cos x(1 - t)} = \lim_{t \rightarrow 1} \frac{\sqrt{2}(1 + t)(1 + t^2)}{t^3 \cos x}$.
As $t \rightarrow 1$,$x \rightarrow \pi/4$,so $\cos x \rightarrow 1/\sqrt{2}$.
$= \frac{\sqrt{2}(1 + 1)(1 + 1^2)}{1^3 \cdot (1/\sqrt{2})} = \frac{\sqrt{2} \cdot 2 \cdot 2}{1/\sqrt{2}} = 4 \cdot 2 = 8$.

(C) Let $L = \lim _{x \rightarrow 0} \frac{2 \tan x+\cos x-1+x}{\sqrt{4 \sin ^2 x+2 \tan x+1}-\sqrt{3 \tan ^2 x+\sin x+1}}$.
Rationalizing the denominator:
$L = \lim _{x \rightarrow 0} \frac{(2 \tan x + \cos x - 1 + x)(\sqrt{4 \sin ^2 x + 2 \tan x + 1} + \sqrt{3 \tan ^2 x + \sin x + 1})}{(4 \sin ^2 x + 2 \tan x + 1) - (3 \tan ^2 x + \sin x + 1)}$.
As $x \rightarrow 0$,the term $(\sqrt{4 \sin ^2 x + 2 \tan x + 1} + \sqrt{3 \tan ^2 x + \sin x + 1}) \rightarrow \sqrt{1} + \sqrt{1} = 2$.
So,$L = 2 \lim _{x \rightarrow 0} \frac{2 \tan x + x + (\cos x - 1)}{4 \sin ^2 x - 3 \tan ^2 x + 2 \tan x - \sin x}$.
Using small angle approximations $\tan x \approx x$,$\sin x \approx x$,$\cos x - 1 \approx -\frac{x^2}{2}$:
$L = 2 \lim _{x \rightarrow 0} \frac{2x + x - \frac{x^2}{2}}{4x^2 - 3x^2 + 2x - x} = 2 \lim _{x \rightarrow 0} \frac{3x - \frac{x^2}{2}}{x^2 + x} = 2 \lim _{x \rightarrow 0} \frac{3 - \frac{x}{2}}{x + 1} = 2 \times \frac{3}{1} = 6$.

(B) We need to evaluate $\lim_{x \to 0^-} f(x)$.
For $x \in (-1, 0)$,the greatest integer function $[x] = -1$.
Therefore,$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{\sin[x]}{[x]}$.
Substituting $[x] = -1$,we get $\frac{\sin(-1)}{-1}$.
Since $\sin(-\theta) = -\sin\theta$,we have $\frac{-\sin 1}{-1} = \sin 1$.

(C) We know that for small $x$,$\sin^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right) = \tan^{-1} x$.
Substituting this into the limit,we get:
$L = \lim_{x \rightarrow 0} \frac{x^4+x^3+x^2}{(\tan^{-1} x)(\tan^{-1} x)} = \lim_{x \rightarrow 0} \frac{x^2(x^2+x+1)}{(\tan^{-1} x)^2}$.
Since $\lim_{x \rightarrow 0} \frac{\tan^{-1} x}{x} = 1$,we can write $(\tan^{-1} x)^2 \approx x^2$ as $x \rightarrow 0$.
$L = \lim_{x \rightarrow 0} \frac{x^2(x^2+x+1)}{x^2} = \lim_{x \rightarrow 0} (x^2+x+1) = 0^2+0+1 = 1$.

(D) Given that,$\lim _{x \rightarrow 0} \frac{1-\cos (1-\cos x)}{\sin ^4 x}$
$= \lim _{x}$ ${\rightarrow 0} \frac{1-\cos (1-\cos x)}{(1-\cos x)^2} \times \left(\frac{1-\cos x}{x^2}\right)^2 \times \left(\frac{x}{\sin x}\right)^4$
As $x \rightarrow 0$,$(1-\cos x) \rightarrow 0$.
Using the standard limit $\lim _{\theta \rightarrow 0} \frac{1-\cos \theta}{\theta^2} = \frac{1}{2}$ and $\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1$:
$= \frac{1}{2} \times \left(\frac{1}{2}\right)^2 \times (1)^4$
$= \frac{1}{2} \times \frac{1}{4} \times 1 = \frac{1}{8}$

(A) Let $L = \lim _{x \rightarrow 0}\left[\tan \left(x+\frac{\pi}{4}\right)\right]^{1 / x}$.
Since this is of the form $1^\infty$,we use the formula $\lim_{x \to a} [f(x)]^{g(x)} = e^{\lim_{x \to a} [f(x)-1]g(x)}$.
$L = e^{\lim _{x \rightarrow 0}\left[\tan \left(x+\frac{\pi}{4}\right)-1\right] \frac{1}{x}}$.
Using the identity $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$,we have $\tan(x+\frac{\pi}{4}) = \frac{\tan x + 1}{1 - \tan x}$.
$L = e^{\lim _{x \rightarrow 0}\left[\frac{\tan x+1}{1-\tan x}-1\right] \frac{1}{x}} = e^{\lim _{x \rightarrow 0} \left[\frac{\tan x+1-1+\tan x}{1-\tan x}\right] \frac{1}{x}}$.
$L = e^{\lim _{x \rightarrow 0} \frac{2 \tan x}{x(1-\tan x)}}$.
Since $\lim_{x \to 0} \frac{\tan x}{x} = 1$ and $\lim_{x \to 0} (1-\tan x) = 1$,we get $L = e^{2 \times 1 / 1} = e^2$.

(C) Given that $f(x) = [x-3] + |x-4|$.
To find $\lim_{x \rightarrow 3^{-}} f(x)$,let $x = 3 - h$,where $h \rightarrow 0$ and $h > 0$.
$\lim_{x \rightarrow 3^{-}} f(x) = \lim_{h \rightarrow 0} ([3 - h - 3] + |3 - h - 4|)$
$= \lim_{h \rightarrow 0} ([-h] + |-1 - h|)$
Since $h$ is a very small positive number,$-h$ is a very small negative number,so $[-h] = -1$.
Also,$|-1 - h| = |-(1 + h)| = 1 + h$.
Therefore,$\lim_{h \rightarrow 0} (-1 + 1 + h) = 0$.

(B) We are given the limit: $\lim _{x \rightarrow 0} \frac{\sin x \sin ^{-1} x}{x^2}$
We can rewrite the expression as: $\lim _{x \rightarrow 0} \left(\frac{\sin x}{x}\right) \left(\frac{\sin ^{-1} x}{x}\right)$
Using the standard limits $\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1$ and $\lim _{x \rightarrow 0} \frac{\sin ^{-1} x}{x} = 1$,we get:
$1 \times 1 = 1$

(B) Given $\lim _{x \rightarrow \infty} x\left(a^{1 / x}+b^{1 / x}+c^{1 / x}-3 k^{1 / x}\right)=0$.
Let $y = \frac{1}{x}$. As $x \rightarrow \infty$,$y \rightarrow 0$.
The expression becomes $\lim _{y \rightarrow 0} \frac{a^y+b^y+c^y-3 k^y}{y} = 0$.
This can be rewritten as $\lim _{y \rightarrow 0} \left[ \frac{a^y-1}{y} + \frac{b^y-1}{y} + \frac{c^y-1}{y} - 3 \frac{k^y-1}{y} \right] = 0$.
Using the standard limit $\lim _{y \rightarrow 0} \frac{a^y-1}{y} = \ln a$,we get:
$\ln a + \ln b + \ln c - 3 \ln k = 0$.
$\ln (abc) = 3 \ln k$.
$\ln (abc) = \ln (k^3)$.
Therefore,$k^3 = abc$,which implies $k = (abc)^{1/3}$.

(C) Let the given expression be $L = \lim _{x}$ ${\rightarrow 0}\left(\frac{4 !}{x^8}\left(1-\cos \frac{x^2}{3}-\cos \frac{x^2}{4}+\cos \frac{x^2}{3} \cos \frac{x^2}{4}\right)\right)$.
We can factor the expression inside the limit as:
$L = \lim _{x \rightarrow 0} \frac{4!}{x^8} (1 - \cos \frac{x^2}{3})(1 - \cos \frac{x^2}{4})$.
Using the identity $1 - \cos \theta = 2 \sin^2(\frac{\theta}{2})$,we get:
$1 - \cos \frac{x^2}{3} = 2 \sin^2(\frac{x^2}{6})$ and $1 - \cos \frac{x^2}{4} = 2 \sin^2(\frac{x^2}{8})$.
Substituting these into the limit:
$L = 24 \times \lim _{x \rightarrow 0} \frac{1}{x^8} \times 2 \sin^2(\frac{x^2}{6}) \times 2 \sin^2(\frac{x^2}{8})$.
$L = 24 \times 4 \times \lim _{x}$ ${\rightarrow 0} \left(\frac{\sin(\frac{x^2}{6})}{\frac{x^2}{6}}\right)^2 \times (\frac{x^2}{6})^2 \times \left(\frac{\sin(\frac{x^2}{8})}{\frac{x^2}{8}}\right)^2 \times (\frac{x^2}{8})^2 \times \frac{1}{x^8}$.
Since $\lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1$,we have:
$L = 96 \times 1^2 \times \frac{x^4}{36} \times 1^2 \times \frac{x^4}{64} \times \frac{1}{x^8}$.
$L = 96 \times \frac{1}{36 \times 64} = \frac{96}{2304} = \frac{1}{24}$.

(B) Let $L = \lim _{x \rightarrow \infty} \left(\frac{2 x^2+3 x+4}{x^2-3 x+5}\right)^{\frac{3|x|+1}{2|x|-1}}$.
As $x \rightarrow \infty$,$|x| = x$.
So,$L = \lim _{x}$ ${\rightarrow \infty} \left(\frac{2 + \frac{3}{x} + \frac{4}{x^2}}{1 - \frac{3}{x} + \frac{5}{x^2}}\right)^{\frac{3 + \frac{1}{x}}{2 - \frac{1}{x}}}$.
Evaluating the limit as $x \rightarrow \infty$,the terms $\frac{1}{x}$ and $\frac{1}{x^2}$ approach $0$.
Thus,$L = \left(\frac{2+0+0}{1-0+0}\right)^{\frac{3+0}{2-0}} = (2)^{\frac{3}{2}}$.
$L = 2^{\frac{3}{2}} = 2^1 \times 2^{\frac{1}{2}} = 2 \sqrt{2}$.

(C) Let $L = \lim _{x \rightarrow 0}\left(\frac{\sinh 2 x}{2 x}\right)^{\frac{1}{x^2}}$.
Using the Taylor series expansion for $\sinh(u) = u + \frac{u^3}{6} + O(u^5)$,we have $\sinh(2x) = 2x + \frac{(2x)^3}{6} + O(x^5) = 2x + \frac{4x^3}{3} + O(x^5)$.
Thus,$\frac{\sinh 2x}{2x} = 1 + \frac{2x^2}{3} + O(x^4)$.
Now,$L = \lim _{x \rightarrow 0} \left(1 + \frac{2x^2}{3} + O(x^4)\right)^{\frac{1}{x^2}}$.
Using the standard limit $\lim _{u \rightarrow 0} (1+u)^{1/u} = e$,let $u = \frac{2x^2}{3}$. As $x \rightarrow 0$,$u \rightarrow 0$.
Then $L = \lim _{x \rightarrow 0} \left[ \left(1 + \frac{2x^2}{3}\right)^{\frac{3}{2x^2}} \right]^{\frac{2}{3}} = e^{2/3}$.

(C) We are given the limit: $\lim _{x \rightarrow \infty} x\left(\log \left(1+\frac{x}{2}\right)-\log \frac{x}{2}\right)$.
Using the property $\log a - \log b = \log \left(\frac{a}{b}\right)$,we have:
$\lim _{x \rightarrow \infty} x \log \left(\frac{1+\frac{x}{2}}{\frac{x}{2}}\right) = \lim _{x \rightarrow \infty} x \log \left(\frac{2}{x} + 1\right)$.
Let $t = \frac{1}{x}$. As $x \rightarrow \infty$,$t \rightarrow 0$. The expression becomes:
$\lim _{t \rightarrow 0} \frac{\log (1+2t)}{t}$.
Using the standard limit $\lim _{u \rightarrow 0} \frac{\log (1+u)}{u} = 1$,we multiply and divide by $2$:
$\lim _{t \rightarrow 0} 2 \cdot \frac{\log (1+2t)}{2t} = 2 \cdot 1 = 2$.

(B) Consider the first limit: $L_1 = \lim _{x \rightarrow-\infty} \frac{3|x|-x}{|x|-2 x}$.
Since $x \rightarrow-\infty$,we have $|x| = -x$.
Substituting this,we get $L_1 = \lim _{x \rightarrow-\infty} \frac{3(-x)-x}{(-x)-2 x} = \lim _{x \rightarrow-\infty} \frac{-4x}{-3x} = \frac{4}{3}$.
Consider the second limit: $L_2 = \lim _{x \rightarrow 0} \frac{\log (1+x^3)}{\sin ^3 x}$.
Using the standard limits $\lim _{u \rightarrow 0} \frac{\log (1+u)}{u} = 1$ and $\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1$,we have:
$L_2 = \lim _{x \rightarrow 0} \left( \frac{\log (1+x^3)}{x^3} \cdot \frac{x^3}{\sin ^3 x} \right) = 1 \cdot 1 = 1$.
Now,subtracting the two results:
$L_1 - L_2 = \frac{4}{3} - 1 = \frac{4-3}{3} = \frac{1}{3}$.

(B) Given $f(x) = \lim_{n \rightarrow \infty} n(x^{1/n} - 1)$. Let $h = \frac{1}{n}$. As $n \rightarrow \infty$,$h \rightarrow 0$.
$f(x) = \lim_{h \rightarrow 0} \frac{x^h - 1}{h}$.
Using the standard limit formula $\lim_{h \rightarrow 0} \frac{a^h - 1}{h} = \ln a$,we get $f(x) = \ln x$.
Now,substitute $f(x) = \ln x$ into the given equation:
$97 f(x) + m f\left(\frac{1}{x}\right) = 0$
$97 \ln x + m \ln\left(\frac{1}{x}\right) = 0$
Since $\ln\left(\frac{1}{x}\right) = -\ln x$,we have:
$97 \ln x - m \ln x = 0$
$(97 - m) \ln x = 0$
For this to hold for all $x > 0$ (where $\ln x \neq 0$),we must have $97 - m = 0$,which implies $m = 97$.

(D) Given,$L = \lim _{x \rightarrow \infty}\left[\frac{x^2+x+3}{x^2-x+2}\right]^x$.
This is of the form $1^\infty$.
Using the formula $\lim _{x \rightarrow \infty}[1+f(x)]^{g(x)} = e^{\lim _{x \rightarrow \infty} f(x) \cdot g(x)}$,we have:
$L = \lim _{x \rightarrow \infty}\left[1+\frac{x^2+x+3}{x^2-x+2}-1\right]^x$
$= \lim _{x \rightarrow \infty}\left[1+\frac{x^2+x+3-x^2+x-2}{x^2-x+2}\right]^x$
$= \lim _{x \rightarrow \infty}\left[1+\frac{2x+1}{x^2-x+2}\right]^x$
$= e^{\lim _{x \rightarrow \infty} \frac{2x+1}{x^2-x+2} \cdot x}$
$= e^{\lim _{x \rightarrow \infty} \frac{2x^2+x}{x^2-x+2}}$
Dividing numerator and denominator by $x^2$:
$= e^{\lim _{x \rightarrow \infty} \frac{2+1/x}{1-1/x+2/x^2}} = e^{\frac{2+0}{1-0+0}} = e^2$.

(C) We use the standard limit formula: $\lim _{x \rightarrow \infty} (1 + \frac{a}{x+b})^{x+c} = e^a$.
Given the expression: $\lim _{x \rightarrow \infty} (\frac{x+6}{x+1})^{x+4}$.
Rewrite the base: $\frac{x+6}{x+1} = \frac{x+1+5}{x+1} = 1 + \frac{5}{x+1}$.
So,the limit becomes $\lim _{x \rightarrow \infty} (1 + \frac{5}{x+1})^{x+4}$.
Using the property $\lim _{x \rightarrow \infty} (1 + \frac{k}{f(x)})^{g(x)} = e^{\lim _{x \rightarrow \infty} k \cdot \frac{g(x)}{f(x)}}$:
$= e^{\lim _{x \rightarrow \infty} 5 \cdot \frac{x+4}{x+1}}$.
$= e^{5 \cdot \lim _{x \rightarrow \infty} \frac{1 + 4/x}{1 + 1/x}} = e^{5 \cdot 1} = e^5$.

(A) We evaluate the limit: $\lim _{x \rightarrow 0} \frac{(1-e^x) \sin x}{x^2+x^3}$
Factor the denominator: $\lim _{x \rightarrow 0} \frac{(1-e^x) \sin x}{x^2(1+x)}$
Rewrite the expression: $\lim _{x}$ ${\rightarrow 0} \left( \frac{1-e^x}{x} \right) \times \left( \frac{\sin x}{x} \right) \times \left( \frac{1}{1+x} \right)$
Using standard limits $\lim _{x \rightarrow 0} \frac{e^x-1}{x} = 1$ and $\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1$:
The expression becomes $(-1) \times (1) \times \left( \frac{1}{1+0} \right) = -1 \times 1 \times 1 = -1$

(B) We evaluate the limit: $\lim _{x \rightarrow 0} \frac{e^x-e^{\sin x}}{2(x-\sin x)}$
By factoring out $e^{\sin x}$ in the numerator,we get:
$= \lim _{x \rightarrow 0} \frac{e^{\sin x}(e^{x-\sin x}-1)}{2(x-\sin x)}$
Using the standard limit $\lim _{u \rightarrow 0} \frac{e^u-1}{u} = 1$,where $u = x - \sin x$:
$= \lim _{x \rightarrow 0} \left( \frac{e^{\sin x}}{2} \cdot \frac{e^{x-\sin x}-1}{x-\sin x} \right)$
As $x \rightarrow 0$,$e^{\sin x} \rightarrow e^0 = 1$ and $\frac{e^{x-\sin x}-1}{x-\sin x} \rightarrow 1$.
Therefore,the limit is $\frac{1}{2} \times 1 = \frac{1}{2}$.

(C) We have $\lim _{n \rightarrow \infty}\left(q^n+p^n\right)^{1 / n}$.
Since $0 < p < q$,we can factor out $q^n$ from the expression:
$= \lim _{n \rightarrow \infty} \left[q^n \left(1 + \left(\frac{p}{q}\right)^n\right)\right]^{1/n}$
$= \lim _{n \rightarrow \infty} q \left(1 + \left(\frac{p}{q}\right)^n\right)^{1/n}$
Since $0 < \frac{p}{q} < 1$,as $n \rightarrow \infty$,$\left(\frac{p}{q}\right)^n \rightarrow 0$.
Therefore,$\lim _{n \rightarrow \infty} q \left(1 + 0\right)^{1/n} = q \cdot 1^0 = q \cdot 1 = q$.

(A) We evaluate the limit $L = \lim _{x \rightarrow \infty} \left(\frac{x+a}{x+b}\right)^{x}$.
This is of the form $1^{\infty}$.
We can rewrite the expression as:
$L = \lim _{x \rightarrow \infty} \left(1 + \frac{a-b}{x+b}\right)^{x}$.
Using the standard limit formula $\lim _{x \rightarrow \infty} (1 + \frac{k}{x})^x = e^k$,we have:
$L = \lim _{x \rightarrow \infty} \left(1 + \frac{a-b}{x+b}\right)^{\frac{x+b}{a-b} \cdot \frac{x(a-b)}{x+b}}$.
$L = e^{\lim _{x \rightarrow \infty} \frac{x(a-b)}{x+b}} = e^{a-b}$.

(C) Let $y = \lim _{n \rightarrow \infty} P\left(1+\frac{r}{100 n}\right)^{t n}$.
We know the standard limit formula $\lim _{k \rightarrow \infty} (1+\frac{x}{k})^k = e^x$.
Here,let $k = n$. Then the expression becomes $P \left[ \lim _{n \rightarrow \infty} (1+\frac{r/100}{n})^n \right]^t$.
Using the limit formula,$\lim _{n \rightarrow \infty} (1+\frac{r/100}{n})^n = e^{r/100}$.
Therefore,$y = P \left( e^{r/100} \right)^t = P e^{\frac{rt}{100}}$.

(C) We know that $\lim _{x \rightarrow \infty} (1 + \frac{a}{x+b})^{x+c} = e^a$.
Given expression: $\lim _{x \rightarrow \infty} (\frac{x+5}{x+2})^{x+3} = \lim _{x \rightarrow \infty} (1 + \frac{3}{x+2})^{x+3}$.
Let $t = x+2$,then as $x \rightarrow \infty$,$t \rightarrow \infty$.
$= \lim _{t \rightarrow \infty} (1 + \frac{3}{t})^{t+1} = \lim _{t \rightarrow \infty} (1 + \frac{3}{t})^t \cdot (1 + \frac{3}{t})^1$.
$= e^3 \cdot 1 = e^3$.

(D) We use the standard limit $\lim _{u \rightarrow 0} \frac{a^u - 1}{u} = \ln a$.
The given expression is $\lim _{x \rightarrow 0} \frac{3^{\sin x}-2^{\tan x}}{\sin x}$.
Dividing the numerator and denominator by $x$,we get:
$\lim _{x \rightarrow 0} \frac{\frac{3^{\sin x}-1}{x} - \frac{2^{\tan x}-1}{x}}{\frac{\sin x}{x}}$.
Since $\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1$ and $\lim _{x \rightarrow 0} \frac{\tan x}{x} = 1$,we have:
$\lim _{x}$ ${\rightarrow 0} \left( \frac{3^{\sin x}-1}{\sin x} \cdot \frac{\sin x}{x} - \frac{2^{\tan x}-1}{\tan x} \cdot \frac{\tan x}{x} \right) / \frac{\sin x}{x}$.
This simplifies to $\ln 3 - \ln 2 = \ln \left( \frac{3}{2} \right)$.

(C) We are given the limit: $\lim _{n \rightarrow \infty} \frac{1}{n^3} \sum_{k=1}^n (k^2 x)$
Since $x$ is independent of $k$,we can write: $\lim _{n \rightarrow \infty} \frac{x}{n^3} \sum_{k=1}^n k^2$
Using the formula for the sum of squares of the first $n$ natural numbers,$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$
Substituting this into the expression: $\lim _{n \rightarrow \infty} \frac{x}{n^3} \cdot \frac{n(n+1)(2n+1)}{6}$
$= \lim _{n \rightarrow \infty} \frac{x}{6} \cdot \frac{n(n+1)(2n+1)}{n^3}$
$= \frac{x}{6} \lim _{n \rightarrow \infty} \frac{n^3(1 + \frac{1}{n})(2 + \frac{1}{n})}{n^3}$
$= \frac{x}{6} \cdot (1 \cdot 2) = \frac{2x}{6} = \frac{x}{3}$

(A) We need to evaluate $\lim _{x \rightarrow 3} \frac{11-[2-x]}{[x+10]}$.
First,consider the left-hand limit $(LHL)$ as $x \rightarrow 3^-$:
As $x \rightarrow 3^-$,$x$ is slightly less than $3$,so $2-x$ is slightly greater than $-1$. Thus,$[2-x] = -1$.
Also,$x+10$ is slightly less than $13$,so $[x+10] = 12$.
$LHL = \lim _{x \rightarrow 3^-} \frac{11-(-1)}{12} = \frac{12}{12} = 1$.
Now,consider the right-hand limit $(RHL)$ as $x \rightarrow 3^+$:
As $x \rightarrow 3^+$,$x$ is slightly greater than $3$,so $2-x$ is slightly less than $-1$. Thus,$[2-x] = -2$.
Also,$x+10$ is slightly greater than $13$,so $[x+10] = 13$.
$RHL = \lim _{x \rightarrow 3^+} \frac{11-(-2)}{13} = \frac{13}{13} = 1$.
Since $LHL = RHL = 1$,the limit exists and is equal to $1$.

(B) The given expression is $\lim _{n \rightarrow \infty} \sum_{k=1}^n \left(\frac{1}{2^k \cdot 3^k} + \frac{1}{2^{k+1} \cdot 3^k}\right)$.
Factor out the common terms in each bracket: $\frac{1}{2^k \cdot 3^k} (1 + \frac{1}{2}) = \frac{1}{6^k} \cdot \frac{3}{2}$.
Thus,the sum becomes $\frac{3}{2} \sum_{k=1}^n \frac{1}{6^k}$.
As $n \rightarrow \infty$,this is an infinite geometric series with first term $a = \frac{1}{6}$ and common ratio $r = \frac{1}{6}$.
The sum of an infinite geometric series is $S = \frac{a}{1-r} = \frac{1/6}{1-1/6} = \frac{1/6}{5/6} = \frac{1}{5}$.
Therefore,the total value is $\frac{3}{2} \times \frac{1}{5} = \frac{3}{10}$.

(D) To find the limit $\lim _{x \rightarrow 0} \frac{\sin |x|}{x}$,we evaluate the left-hand limit $(LHL)$ and the right-hand limit $(RHL)$.
For $x < 0$,$|x| = -x$,so $\lim _{x \rightarrow 0^-} \frac{\sin(-x)}{x} = \lim _{x \rightarrow 0^-} \frac{-\sin x}{x} = -1$.
For $x > 0$,$|x| = x$,so $\lim _{x \rightarrow 0^+} \frac{\sin x}{x} = 1$.
Since the left-hand limit $(-1)$ is not equal to the right-hand limit $(1)$,the limit does not exist.

(B) The function $f(x) = \frac{\sqrt{x+3}}{x+1}$ is defined only when $x+3 \geq 0$,which means $x \geq -3$.
For $x < -3$,the term $\sqrt{x+3}$ involves the square root of a negative number,which is not defined in the set of real numbers.
Since the limit $\lim_{x \rightarrow -3^{-}} f(x)$ requires evaluating the function for values of $x$ slightly less than $-3$,and the function is not defined in that interval,the limit does not exist.

(C) We are given $f(x) = \frac{1}{3} x \sin x - (1 - \cos x)$.
Using the Taylor series expansions for $\sin x$ and $\cos x$ near $x = 0$:
$\sin x = x - \frac{x^3}{6} + O(x^5)$
$\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} + O(x^6)$
Substituting these into $f(x)$:
$f(x) = \frac{1}{3} x \left(x - \frac{x^3}{6} + O(x^5)\right) - \left(1 - (1 - \frac{x^2}{2} + \frac{x^4}{24} + O(x^6))\right)$
$f(x) = \frac{1}{3} x^2 - \frac{x^4}{18} - \frac{x^2}{2} + \frac{x^4}{24} + O(x^6)$
$f(x) = (\frac{1}{3} - \frac{1}{2}) x^2 + (\frac{1}{24} - \frac{1}{18}) x^4 + O(x^6)$
$f(x) = -\frac{1}{6} x^2 - \frac{1}{72} x^4 + O(x^6)$
Now,consider the limit $\lim_{x \rightarrow 0} \frac{f(x)}{x^k} = \lim_{x \rightarrow 0} \frac{-\frac{1}{6} x^2 - \frac{1}{72} x^4 + O(x^6)}{x^k}$.
If $k < 2$,the limit is $\pm \infty$.
If $k = 2$,the limit is $-\frac{1}{6} \neq 0$.
If $k > 2$,the limit is $0$.
Thus,the smallest positive integer $k$ such that the limit is non-zero is $k = 2$.

(A) The interior angle of a regular polygon with $n$ sides is given by the formula: $\theta_n = \frac{(n-2) \times 180^{\circ}}{n}$.
We want to find the limit as $n \rightarrow \infty$:
$\lim_{n \rightarrow \infty} \theta_n = \lim_{n \rightarrow \infty} \frac{(n-2) \times 180^{\circ}}{n} = \lim_{n \rightarrow \infty} (1 - \frac{2}{n}) \times 180^{\circ}$.
As $n \rightarrow \infty$,the term $\frac{2}{n} \rightarrow 0$.
Therefore,the limit is $(1 - 0) \times 180^{\circ} = 180^{\circ} = \pi \text{ radians}$.

(B) The general term is $1 - \frac{1}{\frac{n(n+1)}{2}} = 1 - \frac{2}{n(n+1)} = \frac{n^2+n-2}{n(n+1)} = \frac{(n+2)(n-1)}{n(n+1)}$.
$x_n = \left[ \prod_{k=2}^{n} \frac{(k+2)(k-1)}{k(k+1)} \right]^2$.
$x_n = \left[ \left( \prod_{k=2}^{n} \frac{k+2}{k+1} \right) \left( \prod_{k=2}^{n} \frac{k-1}{k} \right) \right]^2$.
Evaluating the products:
$\prod_{k=2}^{n} \frac{k+2}{k+1} = \frac{4}{3} \cdot \frac{5}{4} \cdot \dots \cdot \frac{n+2}{n+1} = \frac{n+2}{3}$.
$\prod_{k=2}^{n} \frac{k-1}{k} = \frac{1}{2} \cdot \frac{2}{3} \cdot \dots \cdot \frac{n-1}{n} = \frac{1}{n}$.
Thus,$x_n = \left( \frac{n+2}{3} \cdot \frac{1}{n} \right)^2 = \frac{1}{9} \left( \frac{n+2}{n} \right)^2 = \frac{1}{9} \left( 1 + \frac{2}{n} \right)^2$.
Taking the limit as $n \rightarrow \infty$,$\lim_{n \rightarrow \infty} x_n = \frac{1}{9} (1+0)^2 = \frac{1}{9}$.

(B) Let the numerator of the $n^{th}$ term be $a_{n} = 1, 10, 21, 34, 49, \ldots$.
This is a quadratic sequence. The first differences are $9, 11, 13, 15, \ldots$,which form an arithmetic progression.
The general term $a_{n}$ is given by $a_{n} = An^{2} + Bn + C$.
For $n=1, a_{1} = A + B + C = 1$.
For $n=2, a_{2} = 4A + 2B + C = 10$.
For $n=3, a_{3} = 9A + 3B + C = 21$.
Solving these equations: $(a_{2}-a_{1}) = 3A + B = 9$ and $(a_{3}-a_{2}) = 5A + B = 11$.
Subtracting gives $2A = 2 \Rightarrow A = 1$.
Then $3(1) + B = 9 \Rightarrow B = 6$.
Then $1 + 6 + C = 1 \Rightarrow C = -6$.
So,$a_{n} = n^{2} + 6n - 6$.
The $n^{th}$ term of the series is $t_{n} = \frac{n^{2} + 6n - 6}{n!}$.
As $n \rightarrow \infty$,the factorial $n!$ grows much faster than the polynomial $n^{2} + 6n - 6$.
Therefore,$\lim _{n \rightarrow \infty} t_{n} = \lim _{n \rightarrow \infty} \frac{n^{2} + 6n - 6}{n!} = 0$.

(A) We are evaluating $\lim _{x \rightarrow 0}\left\{\frac{1}{x^{2}}+\frac{(2013)^{x}-1}{e^{x}-1}\right\}$.
Using the standard limit $\lim_{x \rightarrow 0} \frac{e^x-1}{x} = 1$,we can rewrite the expression as:
$\lim _{x \rightarrow 0}\left\{\frac{1}{x^{2}}+\frac{(2013)^{x}-1}{x} \cdot \frac{x}{e^{x}-1}\right\}$.
We know that $\lim _{x \rightarrow 0} \frac{(2013)^{x}-1}{x} = \ln(2013)$ and $\lim _{x \rightarrow 0} \frac{x}{e^{x}-1} = 1$.
Thus,the expression becomes $\lim _{x \rightarrow 0} \frac{1}{x^{2}} + \ln(2013) \cdot 1$.
As $x \rightarrow 0$,$\frac{1}{x^2} \rightarrow +\infty$.
Therefore,the limit is $+\infty$.

(C) Let $S = \sum_{n=1}^{1000} (-1)^{n} x^{n} = -x + x^{2} - x^{3} + x^{4} - \dots + x^{1000}$.
This is a finite geometric series with first term $a = -x$,common ratio $r = -x$,and number of terms $n = 1000$.
The sum is given by $S = a \frac{r^{n} - 1}{r - 1} = (-x) \frac{(-x)^{1000} - 1}{-x - 1} = (-x) \frac{x^{1000} - 1}{-(x + 1)} = x \frac{x^{1000} - 1}{x + 1} = \frac{x^{1001} - x}{x + 1}$.
Now,we evaluate the limit as $x \rightarrow \infty$:
$\lim_{x \rightarrow \infty} \frac{x^{1001} - x}{x + 1} = \lim_{x \rightarrow \infty} \frac{x^{1001}(1 - \frac{1}{x^{1000}})}{x(1 + \frac{1}{x})} = \lim_{x \rightarrow \infty} x^{1000} = +\infty$.

(D) Let $L = \lim _{n \rightarrow \infty} \frac{(n !)^{1 / n}}{n} = \lim _{n \rightarrow \infty} \left( \frac{n!}{n^n} \right)^{1/n}$.
Taking the natural logarithm on both sides:
$\ln L = \lim _{n \rightarrow \infty} \frac{1}{n} \ln \left( \frac{n!}{n^n} \right) = \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} \ln \left( \frac{r}{n} \right)$.
This is a Riemann sum for the integral $\int_{0}^{1} \ln(x) \, dx$.
$\ln L = \int_{0}^{1} \ln(x) \, dx = [x \ln x - x]_{0}^{1}$.
Evaluating the limit as $x \rightarrow 0^+$,we get $\lim_{x \rightarrow 0^+} x \ln x = 0$.
So,$\ln L = (1 \ln 1 - 1) - (0) = -1$.
Therefore,$L = e^{-1} = \frac{1}{e}$.

(C) We are given the limit $L = \lim _{x \rightarrow 1} \frac{x+x^2+\ldots+x^n-n}{x-1}$.
Since the form is $\frac{0}{0}$ at $x=1$,we can rewrite the numerator as $(x-1) + (x^2-1) + \ldots + (x^n-1)$.
Thus,$L = \lim _{x \rightarrow 1} \left( \frac{x-1}{x-1} + \frac{x^2-1}{x-1} + \ldots + \frac{x^n-1}{x-1} \right)$.
Using the standard limit formula $\lim _{x \rightarrow a} \frac{x^n-a^n}{x-a} = na^{n-1}$,each term becomes $\lim _{x \rightarrow 1} \frac{x^k-1}{x-1} = k(1)^{k-1} = k$.
Therefore,$L = 1 + 2 + 3 + \ldots + n$.
The sum of the first $n$ natural numbers is $\frac{n(n+1)}{2}$.

(B) We need to evaluate the limit: $\operatorname{Lt}_{x \rightarrow 0} \frac{\sin^2 x + \cos x - 1}{x^2}$.
Using the identity $\sin^2 x = 1 - \cos^2 x$,the expression becomes:
$\operatorname{Lt}_{x \rightarrow 0} \frac{1 - \cos^2 x + \cos x - 1}{x^2} = \operatorname{Lt}_{x \rightarrow 0} \frac{\cos x - \cos^2 x}{x^2}$.
Factor out $\cos x$:
$\operatorname{Lt}_{x \rightarrow 0} \frac{\cos x(1 - \cos x)}{x^2} = \operatorname{Lt}_{x \rightarrow 0} \cos x \cdot \operatorname{Lt}_{x \rightarrow 0} \frac{1 - \cos x}{x^2}$.
We know that $\operatorname{Lt}_{x \rightarrow 0} \cos x = 1$ and $\operatorname{Lt}_{x \rightarrow 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$.
Therefore,the limit is $1 \times \frac{1}{2} = \frac{1}{2}$.

(C) Let $L = \lim _{x \rightarrow 0} \frac{1}{x} \ln \sqrt{\frac{1+x}{1-x}}$.
Using the property $\ln(a^b) = b \ln(a)$,we can rewrite the expression as:
$L = \lim _{x \rightarrow 0} \frac{1}{x} \cdot \frac{1}{2} \ln \left( \frac{1+x}{1-x} \right)$.
$L = \frac{1}{2} \lim _{x \rightarrow 0} \frac{\ln(1+x) - \ln(1-x)}{x}$.
Using the standard limit $\lim _{x \rightarrow 0} \frac{\ln(1+ax)}{x} = a$,we have:
$L = \frac{1}{2} \left( \lim _{x \rightarrow 0} \frac{\ln(1+x)}{x} - \lim _{x \rightarrow 0} \frac{\ln(1-x)}{x} \right)$.
$L = \frac{1}{2} (1 - (-1)) = \frac{1}{2} (2) = 1$.

(C) We evaluate the limit inside the sine function first:
$L = \lim_{x \rightarrow 0} \frac{e^{x}-x-1-\frac{x^{2}}{2}}{x^{2}}$
Using the Taylor series expansion for $e^{x} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \dots$,we get:
$L = \lim_{x \rightarrow 0} \frac{(1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{6} + \dots) - x - 1 - \frac{x^{2}}{2}}{x^{2}}$
$L = \lim_{x \rightarrow 0} \frac{\frac{x^{3}}{6} + \dots}{x^{2}} = \lim_{x \rightarrow 0} (\frac{x}{6} + \dots) = 0$
Since the function $\sin(u)$ is continuous at $u=0$,we have:
$I = \sin(L) = \sin(0) = 0$

(C) We know that $\lim _{x \rightarrow 0} (1+ax)^{1/x} = e^a$.
Using this,$\lim _{x \rightarrow 0} \left(\frac{1+cx}{1-cx}\right)^{1/x} = \frac{\lim _{x \rightarrow 0} (1+cx)^{1/x}}{\lim _{x \rightarrow 0} (1-cx)^{1/x}} = \frac{e^c}{e^{-c}} = e^{2c}$.
Given $e^{2c} = 4$.
Now,we need to find $\lim _{x \rightarrow 0} \left(\frac{1+2cx}{1-2cx}\right)^{1/x} = \frac{\lim _{x \rightarrow 0} (1+2cx)^{1/x}}{\lim _{x \rightarrow 0} (1-2cx)^{1/x}} = \frac{e^{2c}}{e^{-2c}} = e^{4c}$.
Since $e^{2c} = 4$,then $e^{4c} = (e^{2c})^2 = 4^2 = 16$.