Concept of limits, Evaluation of algebric limits Questions in English

(C) We evaluate the limit: $\lim _{x \rightarrow 0} \frac{x \tan 2x - 2x \tan x}{(1 - \cos 2x)^2}$.
Using the identity $1 - \cos 2x = 2 \sin^2 x$,the denominator becomes $(2 \sin^2 x)^2 = 4 \sin^4 x$.
So,the expression is $\lim _{x \rightarrow 0} \frac{x(\tan 2x - 2 \tan x)}{4 \sin^4 x}$.
Using Taylor series expansions: $\tan \theta = \theta + \frac{\theta^3}{3} + \dots$ and $\sin x \approx x$.
$\tan 2x = 2x + \frac{(2x)^3}{3} + O(x^5) = 2x + \frac{8x^3}{3} + O(x^5)$.
$2 \tan x = 2(x + \frac{x^3}{3} + O(x^5)) = 2x + \frac{2x^3}{3} + O(x^5)$.
Numerator: $x[(2x + \frac{8x^3}{3}) - (2x + \frac{2x^3}{3})] = x[\frac{6x^3}{3}] = 2x^4$.
Denominator: $4 \sin^4 x \approx 4x^4$.
Limit: $\lim _{x \rightarrow 0} \frac{2x^4}{4x^4} = \frac{2}{4} = \frac{1}{2}$.

(B) Let $l = \lim _{x}$ ${\rightarrow \frac{\pi}{2}} \left[ \frac{1-\tan \left(\frac{x}{2}\right)}{1+\tan \left(\frac{x}{2}\right)} \right] \left[ \frac{1-\sin x}{(\pi-2 x)^3} \right]$
Since $\tan(\frac{\pi}{4} - \frac{x}{2}) = \frac{1-\tan(x/2)}{1+\tan(x/2)}$,we have:
$l = \lim _{x \rightarrow \frac{\pi}{2}} \frac{\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)(1-\sin x)}{(\pi-2 x)^3}$
Substitute $\pi-2x = \theta$,then $x = \frac{\pi}{2} - \frac{\theta}{2}$. As $x \rightarrow \frac{\pi}{2}$,$\theta \rightarrow 0$.
Also,$\frac{\pi}{4} - \frac{x}{2} = \frac{\pi}{4} - (\frac{\pi}{4} - \frac{\theta}{4}) = \frac{\theta}{4}$ and $1-\sin x = 1-\sin(\frac{\pi}{2}-\frac{\theta}{2}) = 1-\cos(\frac{\theta}{2}) = 2\sin^2(\frac{\theta}{4})$.
Substituting these:
$l = \lim _{\theta \rightarrow 0} \frac{\tan(\frac{\theta}{4}) \cdot 2\sin^2(\frac{\theta}{4})}{\theta^3}$
$l = \lim _{\theta}$ ${\rightarrow 0} \frac{\tan(\frac{\theta}{4})}{\frac{\theta}{4} \cdot 4} \cdot \frac{2\sin^2(\frac{\theta}{4})}{(\frac{\theta}{4})^2 \cdot 16}$
$l = \frac{2}{4 \cdot 16} \lim _{\theta}$ ${\rightarrow 0} \left( \frac{\tan(\frac{\theta}{4})}{\frac{\theta}{4}} \right) \left( \frac{\sin(\frac{\theta}{4})}{\frac{\theta}{4}} \right)^2$
$l = \frac{2}{64} (1)(1)^2 = \frac{1}{32}$

(B) Let $I = \lim _{x \rightarrow \frac{\pi}{2}} \frac{\cot x-\cos x}{(\pi-2 x)^3}$
$= \lim _{x \rightarrow \frac{\pi}{2}} \frac{\cos x(1-\sin x)}{\sin x(\pi-2 x)^3}$
Substitute $x = \frac{\pi}{2}-h$,then $\pi-2x = 2h$. As $x \rightarrow \frac{\pi}{2}$,$h \rightarrow 0$.
$I = \lim _{h \rightarrow 0} \frac{\cos(\frac{\pi}{2}-h)(1-\sin(\frac{\pi}{2}-h))}{\sin(\frac{\pi}{2}-h)(2h)^3}$
$= \lim _{h \rightarrow 0} \frac{\sin h(1-\cos h)}{\cos h \cdot 8h^3}$
Using $1-\cos h = 2\sin^2(\frac{h}{2})$,
$I = \lim _{h \rightarrow 0} \frac{\sin h \cdot 2\sin^2(\frac{h}{2})}{\cos h \cdot 8h^3} = \frac{2}{8} \lim _{h \rightarrow 0} \frac{\sin h}{h} \cdot \lim _{h \rightarrow 0} \frac{\sin^2(\frac{h}{2})}{(\frac{h}{2})^2 \cdot 4} \cdot \lim _{h \rightarrow 0} \frac{1}{\cos h}$
$= \frac{1}{4} \cdot 1 \cdot (1)^2 \cdot \frac{1}{4} \cdot 1 = \frac{1}{16}$

(C) Let $L = \lim _{x \rightarrow 0} \frac{\sqrt{1+x \sin x}-\sqrt{\cos x}}{\tan ^2 \frac{x}{2}}$.
Rationalizing the numerator,we get:
$L = \lim _{x \rightarrow 0} \frac{(1+x \sin x) - \cos x}{\tan ^2 \frac{x}{2} (\sqrt{1+x \sin x} + \sqrt{\cos x})}$.
Using the identity $1 - \cos x = 2 \sin ^2 \frac{x}{2}$,we have:
$L = \lim _{x}$ ${\rightarrow 0} \frac{x \sin x + 2 \sin ^2 \frac{x}{2}}{\tan ^2 \frac{x}{2} (\sqrt{1+x \sin x} + \sqrt{\cos x})}$.
Dividing numerator and denominator by $x^2$:
$L = \lim _{x}$ ${\rightarrow 0} \frac{\frac{\sin x}{x} + 2 \left( \frac{\sin (x/2)}{x} \right)^2}{\left( \frac{\tan (x/2)}{x} \right)^2 (\sqrt{1+x \sin x} + \sqrt{\cos x})}$.
Since $\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1$,$\lim _{x \rightarrow 0} \frac{\sin (x/2)}{x} = \frac{1}{2}$,and $\lim _{x \rightarrow 0} \frac{\tan (x/2)}{x} = \frac{1}{2}$:
$L = \frac{1 + 2(1/2)^2}{(1/2)^2 (\sqrt{1+0} + \sqrt{1})} = \frac{1 + 1/2}{(1/4)(2)} = \frac{3/2}{1/2} = 3$.

(B) We evaluate the limit: $\lim _{x \rightarrow 0} \frac{\sin ^2 x}{\sqrt{2}-\sqrt{1+\cos x}}$
Rationalizing the denominator:
$= \lim _{x \rightarrow 0} \frac{\sin ^2 x (\sqrt{2}+\sqrt{1+\cos x})}{2-(1+\cos x)}$
$= \lim _{x \rightarrow 0} \frac{\sin ^2 x (\sqrt{2}+\sqrt{1+\cos x})}{1-\cos x}$
Using the identity $\sin ^2 x = 1-\cos ^2 x = (1-\cos x)(1+\cos x)$:
$= \lim _{x \rightarrow 0} \frac{(1-\cos x)(1+\cos x)(\sqrt{2}+\sqrt{1+\cos x})}{1-\cos x}$
$= \lim _{x \rightarrow 0} (1+\cos x)(\sqrt{2}+\sqrt{1+\cos x})$
Substituting $x = 0$:
$= (1+\cos 0)(\sqrt{2}+\sqrt{1+\cos 0})$
$= (1+1)(\sqrt{2}+\sqrt{1+1})$
$= 2 \times (\sqrt{2}+\sqrt{2})$
$= 2 \times 2 \sqrt{2} = 4 \sqrt{2}$

(A) We know that $1 - \cos \theta = 2 \sin^2 \left(\frac{\theta}{2}\right)$.
Applying this to the numerator and denominator:
$\lim _{x \rightarrow 0} \frac{\sqrt{2 \sin^2(x^2/2)}}{2 \sin^2(x/2)} = \lim _{x \rightarrow 0} \frac{\sqrt{2} |\sin(x^2/2)|}{2 \sin^2(x/2)}$.
Since $x \rightarrow 0$,$\sin(x^2/2) > 0$,so we can remove the absolute value.
$\lim _{x \rightarrow 0} \frac{\sqrt{2} \sin(x^2/2)}{2 \sin^2(x/2)} = \frac{1}{\sqrt{2}} \lim _{x \rightarrow 0} \frac{\sin(x^2/2)}{x^2/2} \cdot \frac{x^2/2}{\sin^2(x/2)}$.
Using $\lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1$:
$= \frac{1}{\sqrt{2}} \cdot 1 \cdot \lim _{x \rightarrow 0} \frac{x^2/2}{(\sin(x/2))^2} = \frac{1}{\sqrt{2}} \cdot \frac{1}{2} \cdot \lim _{x \rightarrow 0} \left( \frac{x/2}{\sin(x/2)} \right)^2 \cdot 4 = \frac{1}{\sqrt{2}} \cdot \frac{1}{2} \cdot 1^2 \cdot 4 = \frac{2}{\sqrt{2}} = \sqrt{2}$.

(C) We need to evaluate $\lim_{x \rightarrow \infty} \sqrt{\frac{x - \sin x}{x + \cos^{2} x}}$.
Divide the numerator and denominator by $x$ inside the square root:
$= \lim_{x \rightarrow \infty} \sqrt{\frac{1 - \frac{\sin x}{x}}{1 + \frac{\cos^{2} x}{x}}}$
Since $\lim_{x \rightarrow \infty} \frac{\sin x}{x} = 0$ and $\lim_{x \rightarrow \infty} \frac{\cos^{2} x}{x} = 0$ (because $\sin x$ and $\cos^{2} x$ are bounded functions):
$= \sqrt{\frac{1 - 0}{1 + 0}}$
$= \sqrt{1} = 1$

(B) Let $L = \lim _{x \rightarrow 0} \frac{\cos (\sin x)-\cos x}{x^{4}}$.
Using the Taylor series expansion for $\cos \theta = 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - O(\theta^6)$,we have:
$\sin x = x - \frac{x^3}{6} + O(x^5)$
$\cos(\sin x) = 1 - \frac{(x - x^3/6)^2}{2} + \frac{(x - x^3/6)^4}{24} = 1 - \frac{x^2 - x^4/3}{2} + \frac{x^4}{24} = 1 - \frac{x^2}{2} + \frac{x^4}{6} + \frac{x^4}{24} = 1 - \frac{x^2}{2} + \frac{5x^4}{24}$.
$\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24}$.
Substituting these into the limit:
$L = \lim _{x \rightarrow 0} \frac{(1 - \frac{x^2}{2} + \frac{5x^4}{24}) - (1 - \frac{x^2}{2} + \frac{x^4}{24})}{x^4}$.
$L = \lim _{x \rightarrow 0} \frac{\frac{5x^4}{24} - \frac{x^4}{24}}{x^4} = \lim _{x \rightarrow 0} \frac{4x^4/24}{x^4} = \frac{4}{24} = \frac{1}{6}$.

(B) We evaluate the limit: $\lim _{x \rightarrow \infty} x^3 \left(\sqrt{x^2+\sqrt{1+x^4}}-x \sqrt{2}\right)$
Rationalizing the expression:
$= \lim _{x \rightarrow \infty} \frac{x^3 \left(x^2+\sqrt{1+x^4}-2 x^2\right)}{\sqrt{x^2+\sqrt{1+x^4}}+x \sqrt{2}} = \lim _{x \rightarrow \infty} \frac{x^3 \left(\sqrt{1+x^4}-x^2\right)}{\sqrt{x^2+\sqrt{1+x^4}}+x \sqrt{2}}$
Rationalizing the numerator again:
$= \lim _{x \rightarrow \infty} \frac{x^3 \left(1+x^4-x^4\right)}{\left(\sqrt{x^2+\sqrt{1+x^4}}+x \sqrt{2}\right) \left(\sqrt{1+x^4}+x^2\right)}$
Dividing by $x^3$ in the denominator:
$= \lim _{x \rightarrow \infty} \frac{x^3}{x \left(\sqrt{1+\sqrt{\frac{1}{x^4}+1}}+\sqrt{2}\right) \cdot x^2 \left(\sqrt{\frac{1}{x^4}+1}+1\right)}$
$= \lim _{x \rightarrow \infty} \frac{1}{\left(\sqrt{1+\sqrt{\frac{1}{x^4}+1}}+\sqrt{2}\right) \left(\sqrt{\frac{1}{x^4}+1}+1\right)}$
Substituting $x \rightarrow \infty$:
$= \frac{1}{(\sqrt{1+1}+\sqrt{2})(1+1)} = \frac{1}{(\sqrt{2}+\sqrt{2})(2)} = \frac{1}{2 \sqrt{2} \cdot 2} = \frac{1}{4 \sqrt{2}}$

(C) Let $L = \lim _{x \rightarrow 0} \frac{27^x-9^x-3^x+1}{\sqrt{5}-\sqrt{4+\cos x}}$.
Factor the numerator: $27^x-9^x-3^x+1 = 9^x(3^x-1) - 1(3^x-1) = (9^x-1)(3^x-1)$.
Rationalize the denominator: $\frac{1}{\sqrt{5}-\sqrt{4+\cos x}} = \frac{\sqrt{5}+\sqrt{4+\cos x}}{5-(4+\cos x)} = \frac{\sqrt{5}+\sqrt{4+\cos x}}{1-\cos x}$.
So,$L = \lim _{x \rightarrow 0} \frac{(9^x-1)(3^x-1)}{1-\cos x} \cdot (\sqrt{5}+\sqrt{4+\cos x})$.
Using $1-\cos x = 2 \sin^2(x/2)$,we have $L = \lim _{x \rightarrow 0} \frac{(9^x-1)(3^x-1)}{2 \sin^2(x/2)} \cdot (\sqrt{5}+\sqrt{4+\cos x})$.
Multiply and divide by $x^2$: $L = \lim _{x}$ ${\rightarrow 0} \frac{(\frac{9^x-1}{x})(\frac{3^x-1}{x})}{2 \cdot (\frac{\sin(x/2)}{x/2})^2 \cdot (1/4)} \cdot (\sqrt{5}+\sqrt{4+\cos x})$.
$L = \frac{\ln 9 \cdot \ln 3}{2 \cdot 1 \cdot (1/4)} \cdot (\sqrt{5}+\sqrt{5}) = \frac{2 \ln 3 \cdot \ln 3}{1/2} \cdot 2 \sqrt{5} = 8 \sqrt{5} (\ln 3)^2$.

(B) Let $L = \lim _{x \rightarrow 1} \frac{2^{2 x-2}-2^x+1}{\sin ^2(x-1)}$.
Note that $2^{2x-2} - 2^x + 1$ can be written as $(2^{x-1})^2 - 2 \cdot 2^{x-1} + 1$ is incorrect,let's re-evaluate the numerator: $2^{2x-2} - 2^x + 1 = (2^{x-1})^2 - 2 \cdot 2^{x-1} + 1 = (2^{x-1}-1)^2$.
So,$L = \lim _{x \rightarrow 1} \frac{(2^{x-1}-1)^2}{\sin ^2(x-1)}$.
Divide numerator and denominator by $(x-1)^2$:
$L = \lim _{x \rightarrow 1} \frac{\left(\frac{2^{x-1}-1}{x-1}\right)^2}{\left(\frac{\sin(x-1)}{x-1}\right)^2}$.
Using the standard limits $\lim _{h \rightarrow 0} \frac{a^h-1}{h} = \log a$ and $\lim _{h \rightarrow 0} \frac{\sin h}{h} = 1$,where $h = x-1$:
$L = \frac{(\log 2)^2}{1^2} = (\log 2)^2$.

(A) Given limit: $\lim _{x \rightarrow 0} \frac{15^{x}-5^{x}-3^{x}+1}{1-\cos 2 x}$
Factorizing the numerator: $15^{x}-5^{x}-3^{x}+1 = 5^{x}(3^{x}-1) - 1(3^{x}-1) = (3^{x}-1)(5^{x}-1)$
Using the identity $1-\cos 2x = 2\sin^{2}x$:
$\lim _{x \rightarrow 0} \frac{(3^{x}-1)(5^{x}-1)}{2\sin^{2}x}$
Dividing numerator and denominator by $x^{2}$:
$\lim _{x \rightarrow 0} \frac{(\frac{3^{x}-1}{x})(\frac{5^{x}-1}{x})}{2(\frac{\sin x}{x})^{2}}$
Using the standard limit $\lim _{x \rightarrow 0} \frac{a^{x}-1}{x} = \log a$ and $\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1$:
$= \frac{(\log 3)(\log 5)}{2(1)^{2}} = \frac{(\log 3)(\log 5)}{2}$

(B) Let $L = \lim _{x \rightarrow 1} (1 + \log _{e} x)^{1 / \log _{e} x}$.
As $x \rightarrow 1$,$\log _{e} x \rightarrow 0$,so the expression takes the form $1^{\infty}$.
We use the standard limit formula: $\lim _{u \rightarrow 0} (1 + u)^{1/u} = e$.
Let $u = \log _{e} x$. As $x \rightarrow 1$,$u \rightarrow 0$.
Substituting this into the limit,we get $\lim _{u \rightarrow 0} (1 + u)^{1/u} = e$.
Therefore,the correct option is $B$.

(D) We know that $\lim _{x \rightarrow \infty} (1 + \frac{a}{x})^x = e^a$.
Given expression is $L = \lim _{x \rightarrow \infty} (\frac{x+8}{x+1})^{x+5}$.
Rewrite the base: $\frac{x+8}{x+1} = \frac{x+1+7}{x+1} = 1 + \frac{7}{x+1}$.
So,$L = \lim _{x \rightarrow \infty} (1 + \frac{7}{x+1})^{x+5}$.
Let $u = x+1$,then as $x \rightarrow \infty$,$u \rightarrow \infty$ and $x = u-1$.
$L = \lim _{u \rightarrow \infty} (1 + \frac{7}{u})^{u-1+5} = \lim _{u \rightarrow \infty} (1 + \frac{7}{u})^{u+4}$.
$L = \lim _{u \rightarrow \infty} (1 + \frac{7}{u})^u \cdot \lim _{u \rightarrow \infty} (1 + \frac{7}{u})^4$.
$L = e^7 \cdot (1 + 0)^4 = e^7 \cdot 1 = e^7$.

(C) Given $A = \lim_{x \rightarrow 0^{+}} \left(1 + \tan^2 \sqrt{x}\right)^{\frac{1}{2x}}$.
This is of the form $1^{\infty}$.
Using the formula $\lim_{x \rightarrow a} f(x)^{g(x)} = e^{\lim_{x \rightarrow a} (f(x)-1)g(x)}$,we have:
$\log_{e} A = \lim_{x \rightarrow 0^{+}} \left(1 + \tan^2 \sqrt{x} - 1\right) \cdot \frac{1}{2x}$
$= \lim_{x \rightarrow 0^{+}} \frac{\tan^2 \sqrt{x}}{2x}$
$= \lim_{x \rightarrow 0^{+}} \frac{1}{2} \left(\frac{\tan \sqrt{x}}{\sqrt{x}}\right)^2$
Since $\lim_{\theta \rightarrow 0} \frac{\tan \theta}{\theta} = 1$,we get:
$\log_{e} A = \frac{1}{2} \cdot (1)^2 = \frac{1}{2}$.

(B) Let $L = \lim _{x \rightarrow 0}\left(\frac{1+\tan x}{1+\sin x}\right)^{\operatorname{cosec} x}$.
Since the form is $1^\infty$,we use the formula $\lim_{x \to a} f(x)^{g(x)} = e^{\lim_{x \to a} (f(x)-1)g(x)}$.
$L = e^{\lim _{x \rightarrow 0}\left(\frac{1+\tan x}{1+\sin x}-1\right) \operatorname{cosec} x}$
$L = e^{\lim _{x \rightarrow 0}\left(\frac{\tan x-\sin x}{1+\sin x}\right) \cdot \frac{1}{\sin x}}$
$L = e^{\lim _{x \rightarrow 0}\left(\frac{\frac{\sin x}{\cos x}-\sin x}{1+\sin x}\right) \cdot \frac{1}{\sin x}}$
$L = e^{\lim _{x \rightarrow 0}\left(\frac{\sin x(1-\cos x)}{\cos x(1+\sin x)}\right) \cdot \frac{1}{\sin x}}$
$L = e^{\lim _{x \rightarrow 0}\frac{1-\cos x}{\cos x(1+\sin x)}}$
As $x \to 0$,$\cos x \to 1$ and $1-\cos x \to 0$.
$L = e^{\frac{0}{1(1+0)}} = e^0 = 1$.

(D) The limit is of the form $1^\infty$. We use the formula $\lim_{x \to a} f(x)^{g(x)} = e^{\lim_{x \to a} (f(x)-1)g(x)}$.
$\lim _{x \rightarrow 2}\left(\frac{5 x-8}{8-3 x}\right)^{\frac{3}{2 x-4}} = e^{\lim _{x \rightarrow 2}\left(\frac{5 x-8}{8-3 x}-1\right) \times \frac{3}{2 x-4}}$
$= e^{\lim _{x \rightarrow 2}\left(\frac{5 x-8 - (8-3 x)}{8-3 x}\right) \times \frac{3}{2(x-2)}}$
$= e^{\lim _{x \rightarrow 2}\left(\frac{8x-16}{8-3 x}\right) \times \frac{3}{2(x-2)}}$
$= e^{\lim _{x \rightarrow 2}\left(\frac{8(x-2)}{8-3 x}\right) \times \frac{3}{2(x-2)}}$
$= e^{\lim _{x \rightarrow 2} \frac{24}{2(8-3 x)}}$
$= e^{\frac{12}{8-3(2)}} = e^{\frac{12}{2}} = e^6$

(A) Let $L = \lim_{x \to \infty} \left(1 + \frac{1}{a+bx}\right)^{c+dx}$. This is in the form of $1^{\infty}$.
Using the formula $\lim_{x \to \infty} (1 + f(x))^{g(x)} = e^{\lim_{x \to \infty} f(x)g(x)}$:
$L = e^{\lim_{x \to \infty} \left(\frac{1}{a+bx}\right) \cdot (c+dx)}$
$L = e^{\lim_{x \to \infty} \frac{c+dx}{a+bx}}$
Divide the numerator and denominator by $x$:
$L = e^{\lim_{x \to \infty} \frac{c/x + d}{a/x + b}}$
As $x \to \infty$,$c/x \to 0$ and $a/x \to 0$:
$L = e^{\frac{0+d}{0+b}} = e^{d/b}$

(A) The given limit is of the form $1^{\infty}$.
We use the formula $\lim _{x \rightarrow a} [f(x)]^{g(x)} = e^{\lim _{x \rightarrow a} g(x)[f(x)-1]}$.
$\lim _{x \rightarrow \infty}\left(\frac{x^{2}-2 x+1}{x^{2}-4 x+2}\right)^{x} = e^{\lim _{x \rightarrow \infty} x \left(\frac{x^{2}-2 x+1}{x^{2}-4 x+2} - 1\right)}$
$= e^{\lim _{x \rightarrow \infty} x \left(\frac{x^{2}-2 x+1 - (x^{2}-4 x+2)}{x^{2}-4 x+2}\right)}$
$= e^{\lim _{x \rightarrow \infty} x \left(\frac{2 x-1}{x^{2}-4 x+2}\right)}$
$= e^{\lim _{x \rightarrow \infty} \frac{2 x^{2}-x}{x^{2}-4 x+2}}$
Dividing numerator and denominator by $x^{2}$,we get:
$= e^{\lim _{x \rightarrow \infty} \frac{2 - 1/x}{1 - 4/x + 2/x^{2}}} = e^{2/1} = e^{2}$.

(D) The given expression is $S_n = \sum_{k=1}^{n} \frac{k^3}{1-n^4}$.
Since the denominator $(1-n^4)$ is independent of the summation index $k$,we can write:
$S_n = \frac{1}{1-n^4} \sum_{k=1}^{n} k^3$.
Using the formula for the sum of cubes,$\sum_{k=1}^{n} k^3 = \left[\frac{n(n+1)}{2}\right]^2 = \frac{n^2(n+1)^2}{4}$.
Thus,$S_n = \frac{n^2(n+1)^2}{4(1-n^4)}$.
Expanding the numerator: $S_n = \frac{n^2(n^2+2n+1)}{4(1-n^4)} = \frac{n^4+2n^3+n^2}{4-4n^4}$.
Now,taking the limit as $n \rightarrow \infty$:
$\lim _{n \rightarrow \infty} S_n = \lim _{n \rightarrow \infty} \frac{n^4+2n^3+n^2}{4-4n^4}$.
Dividing numerator and denominator by $n^4$:
$\lim _{n \rightarrow \infty} \frac{1+\frac{2}{n}+\frac{1}{n^2}}{\frac{4}{n^4}-4} = \frac{1+0+0}{0-4} = -\frac{1}{4}$.

(B) The given expression is $S_n = \sum_{k=1}^{n} \frac{k^2}{n^3+1}$.
Since the denominator $n^3+1$ is independent of $k$,we can write $S_n = \frac{1}{n^3+1} \sum_{k=1}^{n} k^2$.
Using the formula for the sum of squares of the first $n$ natural numbers,$\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$.
Thus,$S_n = \frac{n(n+1)(2n+1)}{6(n^3+1)}$.
To find the limit as $n \rightarrow \infty$,we evaluate $\lim _{n \rightarrow \infty} \frac{2n^3+3n^2+n}{6n^3+6}$.
Dividing the numerator and denominator by $n^3$,we get $\lim _{n \rightarrow \infty} \frac{2 + \frac{3}{n} + \frac{1}{n^2}}{6 + \frac{6}{n^3}} = \frac{2}{6} = \frac{1}{3}$.

(C) We know the formulas for the sums of powers of the first $n$ natural numbers:
$S_1 = \frac{n(n+1)}{2}$,$S_2 = \frac{n(n+1)(2n+1)}{6}$,$S_3 = \frac{n^2(n+1)^2}{4}$.
Given the expression $L = \lim_{n \rightarrow \infty} \frac{S_1(1 + \frac{S_3}{4})}{S_2^2}$.
Substituting the formulas:
$L = \lim_{n \rightarrow \infty} \frac{\frac{n(n+1)}{2} (1 + \frac{n^2(n+1)^2}{16})}{\frac{n^2(n+1)^2(2n+1)^2}{36}}$.
Simplifying the expression:
$L = \lim_{n \rightarrow \infty} \frac{n(n+1)}{2} \cdot \frac{16 + n^2(n+1)^2}{16} \cdot \frac{36}{n^2(n+1)^2(2n+1)^2}$.
$L = \lim_{n \rightarrow \infty} \frac{18}{16} \cdot \frac{n(n+1) [16 + n^2(n+1)^2]}{n^2(n+1)^2(2n+1)^2}$.
$L = \frac{9}{8} \lim_{n \rightarrow \infty} \frac{16 + n^2(n+1)^2}{n(n+1)(2n+1)^2}$.
Dividing the numerator and denominator by $n^4$:
$L = \frac{9}{8} \lim_{n}$ ${\rightarrow \infty} \frac{\frac{16}{n^4} + (1 + \frac{1}{n})^2}{(1 + \frac{1}{n}) (2 + \frac{1}{n})^2} = \frac{9}{8} \cdot \frac{0 + 1}{1 \cdot 4} = \frac{9}{32}$.

(D) Let $f(x) = \frac{x}{|x| + x^2}$.
For the left-hand limit as $x \rightarrow 0^-$,we have $|x| = -x$:
$\lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0} \frac{x}{-x + x^2} = \lim_{x \rightarrow 0} \frac{1}{-1 + x} = -1$.
For the right-hand limit as $x \rightarrow 0^+$,we have $|x| = x$:
$\lim_{x \rightarrow 0^+} f(x) = \lim_{x \rightarrow 0} \frac{x}{x + x^2} = \lim_{x \rightarrow 0} \frac{1}{1 + x} = 1$.
Since $\lim_{x \rightarrow 0^-} f(x) \neq \lim_{x \rightarrow 0^+} f(x)$,the limit does not exist.

(B) First,we evaluate $a$:
$a = \lim_{n \rightarrow \infty} \frac{n(n+1)}{2n^2} = \lim_{n \rightarrow \infty} \frac{n^2+n}{2n^2} = \lim_{n \rightarrow \infty} \frac{1 + \frac{1}{n}}{2} = \frac{1}{2}$.
Next,we evaluate $b$:
$b = \lim_{n \rightarrow \infty} \frac{n(n+1)(2n+1)}{6n^3} = \lim_{n \rightarrow \infty} \frac{2n^3 + 3n^2 + n}{6n^3} = \lim_{n \rightarrow \infty} \frac{2 + \frac{3}{n} + \frac{1}{n^2}}{6} = \frac{2}{6} = \frac{1}{3}$.
Comparing the values,$a = \frac{1}{2}$ and $b = \frac{1}{3}$.
Therefore,$2a = 2(\frac{1}{2}) = 1$ and $3b = 3(\frac{1}{3}) = 1$.
Thus,$2a = 3b$.

(B) Given $f(x) = \begin{cases} x^2-1, & 0 < x < 2 \\ 2x+3, & 2 \leq x < 3 \end{cases}$.
First,we find the left-hand limit at $x=2$:
$\lim_{x \rightarrow 2^{-}} f(x) = \lim_{x \rightarrow 2^{-}} (x^2-1) = 2^2 - 1 = 3$.
Next,we find the right-hand limit at $x=2$:
$\lim_{x \rightarrow 2^{+}} f(x) = \lim_{x \rightarrow 2^{+}} (2x+3) = 2(2) + 3 = 7$.
The roots of the required quadratic equation are $\alpha = 3$ and $\beta = 7$.
The quadratic equation is given by $x^2 - (\alpha + \beta)x + (\alpha \times \beta) = 0$.
Substituting the values: $x^2 - (3 + 7)x + (3 \times 7) = 0$.
Thus,the equation is $x^2 - 10x + 21 = 0$.

(D) We are given the limit $\lim_{x \rightarrow 0} \frac{|x|}{x}$.
To evaluate this,we check the left-hand limit and the right-hand limit.
For the right-hand limit $(x \rightarrow 0^{+})$,$|x| = x$,so $\lim_{x \rightarrow 0^{+}} \frac{x}{x} = 1$.
For the left-hand limit $(x \rightarrow 0^{-})$,$|x| = -x$,so $\lim_{x \rightarrow 0^{-}} \frac{-x}{x} = -1$.
Since the left-hand limit $\neq$ the right-hand limit,the limit does not exist.

(D) We use the trigonometric identity $\sin C - \sin D = 2 \cos \left(\frac{C+D}{2}\right) \sin \left(\frac{C-D}{2}\right)$.
Given the limit $\lim _{x \rightarrow 0} \frac{\sin (2+x)-\sin (2-x)}{x} = A \cos B$.
Applying the identity with $C = 2+x$ and $D = 2-x$:
$\lim _{x \rightarrow 0} \frac{2 \cos \left(\frac{2+x+2-x}{2}\right) \sin \left(\frac{2+x-(2-x)}{2}\right)}{x} = A \cos B$.
$\lim _{x \rightarrow 0} \frac{2 \cos (2) \sin (x)}{x} = A \cos B$.
Since $\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1$,we get:
$2 \cos 2 = A \cos B$.
Comparing both sides,we find $A = 2$ and $B = 2$.

(A) Let $L = \lim _{y \rightarrow 0} \frac{\sqrt{3+y^3}-\sqrt{3}}{y^3}$
Rationalizing the numerator:
$L = \lim _{y}$ ${\rightarrow 0} \frac{(\sqrt{3+y^3}-\sqrt{3})}{y^3} \times \frac{(\sqrt{3+y^3}+\sqrt{3})}{(\sqrt{3+y^3}+\sqrt{3})}$
$L = \lim _{y \rightarrow 0} \frac{3+y^3-3}{y^3(\sqrt{3+y^3}+\sqrt{3})}$
$L = \lim _{y \rightarrow 0} \frac{y^3}{y^3(\sqrt{3+y^3}+\sqrt{3})}$
$L = \lim _{y \rightarrow 0} \frac{1}{\sqrt{3+y^3}+\sqrt{3}}$
Substituting $y = 0$:
$L = \frac{1}{\sqrt{3+0}+\sqrt{3}} = \frac{1}{\sqrt{3}+\sqrt{3}} = \frac{1}{2 \sqrt{3}}$

(B) For Statement $1$:
$\lim _{x \rightarrow 1} \frac{a x^{2}+b x+c}{c x^{2}+b x+a} = \frac{a(1)^{2}+b(1)+c}{c(1)^{2}+b(1)+a} = \frac{a+b+c}{a+b+c} = 1$.
Since $a+b+c \neq 0$,the limit is $1$. Thus,Statement $1$ is true.
For Statement $2$:
$\lim _{x \rightarrow -2} \frac{\frac{1}{x}+\frac{1}{2}}{x+2} = \lim _{x \rightarrow -2} \frac{\frac{2+x}{2x}}{x+2} = \lim _{x \rightarrow -2} \frac{2+x}{2x(x+2)} = \lim _{x \rightarrow -2} \frac{1}{2x} = \frac{1}{2(-2)} = -\frac{1}{4}$.
Since $-\frac{1}{4} \neq \frac{1}{4}$,Statement $2$ is false.

(C) We are given the limit: $\lim _{x \rightarrow 0} \frac{x e^{x}-\sin x}{x}$.
By splitting the fraction,we get:
$\lim _{x \rightarrow 0} \left( \frac{x e^{x}}{x} - \frac{\sin x}{x} \right)$
$= \lim _{x \rightarrow 0} e^{x} - \lim _{x \rightarrow 0} \frac{\sin x}{x}$
Using the standard limit $\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1$ and evaluating $e^{0} = 1$:
$= 1 - 1 = 0$.

(D) To evaluate the limit $L = \lim _{x \rightarrow a} \frac{\sqrt{a+2x} - \sqrt{3x}}{\sqrt{3a+x} - 2\sqrt{x}}$,we rationalize the numerator and the denominator:
$L = \lim _{x}$ ${\rightarrow a} \left( \frac{\sqrt{a+2x} - \sqrt{3x}}{\sqrt{a+2x} + \sqrt{3x}} \cdot \frac{\sqrt{3a+x} + 2\sqrt{x}}{\sqrt{3a+x} - 2\sqrt{x}} \cdot \frac{\sqrt{3a+x} + 2\sqrt{x}}{\sqrt{a+2x} + \sqrt{3x}} \right)$
$L = \lim _{x}$ ${\rightarrow a} \left( \frac{(a+2x) - 3x}{(3a+x) - 4x} \cdot \frac{\sqrt{3a+x} + 2\sqrt{x}}{\sqrt{a+2x} + \sqrt{3x}} \right)$
$L = \lim _{x}$ ${\rightarrow a} \left( \frac{a-x}{3a-3x} \cdot \frac{\sqrt{3a+x} + 2\sqrt{x}}{\sqrt{a+2x} + \sqrt{3x}} \right)$
$L = \lim _{x}$ ${\rightarrow a} \left( \frac{a-x}{3(a-x)} \cdot \frac{\sqrt{3a+x} + 2\sqrt{x}}{\sqrt{a+2x} + \sqrt{3x}} \right)$
$L = \frac{1}{3} \cdot \frac{\sqrt{3a+a} + 2\sqrt{a}}{\sqrt{a+2a} + \sqrt{3a}} = \frac{1}{3} \cdot \frac{\sqrt{4a} + 2\sqrt{a}}{\sqrt{3a} + \sqrt{3a}}$
$L = \frac{1}{3} \cdot \frac{2\sqrt{a} + 2\sqrt{a}}{2\sqrt{3a}} = \frac{1}{3} \cdot \frac{4\sqrt{a}}{2\sqrt{3}\sqrt{a}} = \frac{2}{3\sqrt{3}}$

(B) Let $L = \lim _{n \rightarrow \infty} n \sin \frac{2 \pi}{3 n} \cos \frac{2 \pi}{3 n}$.
We know that $\sin(2\theta) = 2 \sin \theta \cos \theta$,so $\sin \theta \cos \theta = \frac{1}{2} \sin(2\theta)$.
Here,$\theta = \frac{2 \pi}{3 n}$.
Thus,$L = \lim _{n \rightarrow \infty} n \cdot \frac{1}{2} \sin \left( 2 \cdot \frac{2 \pi}{3 n} \right) = \lim _{n \rightarrow \infty} \frac{n}{2} \sin \left( \frac{4 \pi}{3 n} \right)$.
Using the standard limit $\lim _{x \rightarrow 0} \frac{\sin x}{x} = 1$,we rewrite the expression:
$L = \lim _{n \rightarrow \infty} \frac{n}{2} \cdot \left( \frac{4 \pi}{3 n} \right) \cdot \frac{\sin \left( \frac{4 \pi}{3 n} \right)}{\left( \frac{4 \pi}{3 n} \right)}$.
As $n \rightarrow \infty$,$\frac{4 \pi}{3 n} \rightarrow 0$,so $\frac{\sin \left( \frac{4 \pi}{3 n} \right)}{\left( \frac{4 \pi}{3 n} \right)} \rightarrow 1$.
Therefore,$L = \lim _{n \rightarrow \infty} \frac{n}{2} \cdot \frac{4 \pi}{3 n} \cdot 1 = \frac{4 \pi}{6} = \frac{2 \pi}{3}$.

(C) To evaluate the limit $\lim _{n \rightarrow \infty} \frac{3 \cdot 2^{n+1}-4 \cdot 5^{n+1}}{5 \cdot 2^{n}+7 \cdot 5^{n}}$,we divide the numerator and the denominator by $5^n$:
$\lim _{n \rightarrow \infty} \frac{3 \cdot 2 \cdot 2^{n} - 4 \cdot 5 \cdot 5^{n}}{5 \cdot 2^{n} + 7 \cdot 5^{n}}$
$= \lim _{n \rightarrow \infty} \frac{6 \cdot 2^{n} - 20 \cdot 5^{n}}{5 \cdot 2^{n} + 7 \cdot 5^{n}}$
Dividing by $5^n$:
$= \lim _{n \rightarrow \infty} \frac{6 \cdot (\frac{2}{5})^{n} - 20}{5 \cdot (\frac{2}{5})^{n} + 7}$
Since $\lim _{n \rightarrow \infty} (\frac{2}{5})^{n} = 0$,we get:
$= \frac{6 \cdot 0 - 20}{5 \cdot 0 + 7} = -\frac{20}{7}$

(A) Given that $x = \sum_{r=1}^{n}(2r-1)$.
This is the sum of the first $n$ odd integers,which is $x = n^2$.
Therefore,$x^2 = n^4$.
The expression becomes $\lim_{n \rightarrow \infty} \left[ \frac{1^3 + 2^3 + \ldots + n^3}{x^2} \right]$.
Using the formula for the sum of cubes,$\sum_{k=1}^{n} k^3 = \frac{n^2(n+1)^2}{4}$.
Substituting this into the limit: $\lim_{n \rightarrow \infty} \left[ \frac{n^2(n+1)^2}{4n^4} \right]$.
$= \lim_{n \rightarrow \infty} \left[ \frac{n^2 \cdot n^2(1 + \frac{1}{n})^2}{4n^4} \right]$.
$= \lim_{n \rightarrow \infty} \left[ \frac{(1 + \frac{1}{n})^2}{4} \right] = \frac{1}{4}$.

(B) Given $f(x) = \begin{cases} \frac{e^{1 / x}-1}{e^{1 / x}+1}, & x \neq 0 \\ 0, & x=0 \end{cases}$
Right hand limit $(RHL)$:
$\lim_{x \rightarrow 0^{+}} f(x) = \lim_{h \rightarrow 0} f(0+h) = \lim_{h \rightarrow 0} \frac{e^{1/h}-1}{e^{1/h}+1}$
Dividing numerator and denominator by $e^{1/h}$:
$\lim_{h \rightarrow 0} \frac{1 - e^{-1/h}}{1 + e^{-1/h}} = \frac{1 - 0}{1 + 0} = 1$
Left hand limit $(LHL)$:
$\lim_{x \rightarrow 0^{-}} f(x) = \lim_{h \rightarrow 0} f(0-h) = \lim_{h \rightarrow 0} \frac{e^{-1/h}-1}{e^{-1/h}+1}$
As $h \rightarrow 0^{+}$,$e^{-1/h} \rightarrow 0$:
$\frac{0 - 1}{0 + 1} = -1$
Thus,the $RHL$ is $1$ and the $LHL$ is $-1$.

(A) We have,$\lim _{x \rightarrow 0} \left( \frac{\tan x}{\sqrt{2x+4}-2} \right)$.
To evaluate this limit,we rationalize the denominator:
$= \lim _{x \rightarrow 0} \frac{\tan x (\sqrt{2x+4}+2)}{(\sqrt{2x+4}-2)(\sqrt{2x+4}+2)}$
$= \lim _{x \rightarrow 0} \frac{\tan x (\sqrt{2x+4}+2)}{(2x+4)-4}$
$= \lim _{x \rightarrow 0} \frac{\tan x (\sqrt{2x+4}+2)}{2x}$
$= \lim _{x \rightarrow 0} \left( \frac{\tan x}{x} \right) \times \frac{\sqrt{2x+4}+2}{2}$
Using the standard limit $\lim _{x \rightarrow 0} \frac{\tan x}{x} = 1$:
$= 1 \times \frac{\sqrt{2(0)+4}+2}{2}$
$= \frac{\sqrt{4}+2}{2} = \frac{2+2}{2} = \frac{4}{2} = 2$.

(B) First,we calculate $l$:
$l = \lim_{\theta \rightarrow 0} \left( \frac{3 \sin \theta - 4 \sin^2 \theta}{\theta} \right) = \lim_{\theta \rightarrow 0} \left( 3 \frac{\sin \theta}{\theta} - 4 \sin \theta \cdot \frac{\sin \theta}{\theta} \right)$
Since $\lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1$,we have $l = 3(1) - 4(0)(1) = 3$.
Next,we calculate $m$:
$m = \lim_{\theta \rightarrow 0} \frac{2 \tan \theta}{\theta(1 - \tan^2 \theta)} = \lim_{\theta \rightarrow 0} \frac{\tan(2\theta)}{\theta}$
Using the limit $\lim_{x \rightarrow 0} \frac{\tan(kx)}{x} = k$,we get $m = 2$.
The quadratic equation with roots $l=3$ and $m=2$ is given by $x^2 - (l+m)x + lm = 0$.
Substituting the values,we get $x^2 - (3+2)x + (3 \times 2) = 0$,which simplifies to $x^2 - 5x + 6 = 0$.

(B) The given expression is $\lim _{n \rightarrow \infty} \frac{2^2+4^2+6^2+\ldots+(2 n)^2}{n^3}$.
We can factor out $2^2 = 4$ from the numerator:
$= \lim _{n \rightarrow \infty} \frac{4(1^2+2^2+3^2+\ldots+n^2)}{n^3}$.
Using the formula for the sum of squares of the first $n$ natural numbers,$\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$:
$= 4 \lim _{n \rightarrow \infty} \frac{n(n+1)(2n+1)}{6n^3}$.
$= \frac{4}{6} \lim _{n \rightarrow \infty} \frac{n \cdot n(1+\frac{1}{n}) \cdot n(2+\frac{1}{n})}{n^3}$.
$= \frac{2}{3} \lim _{n \rightarrow \infty} (1+\frac{1}{n})(2+\frac{1}{n})$.
As $n \rightarrow \infty$,$\frac{1}{n} \rightarrow 0$,so the limit is $\frac{2}{3} \times (1)(2) = \frac{4}{3}$.
Thus,the correct option is $B$.

(D) We are given the limit: $\lim _{x \rightarrow 0} \frac{\tan x+4 \tan 2 x-3 \tan 3 x}{x^2 \tan x}$.
Using the expansion $\tan x = x + \frac{x^3}{3} + O(x^5)$,we have:
$\tan x = x + \frac{x^3}{3} + O(x^5)$
$\tan 2x = 2x + \frac{(2x)^3}{3} + O(x^5) = 2x + \frac{8x^3}{3} + O(x^5)$
$\tan 3x = 3x + \frac{(3x)^3}{3} + O(x^5) = 3x + 9x^3 + O(x^5)$
Substituting these into the numerator:
Numerator $= (x + \frac{x^3}{3}) + 4(2x + \frac{8x^3}{3}) - 3(3x + 9x^3) + O(x^5)$
$= x + \frac{x^3}{3} + 8x + \frac{32x^3}{3} - 9x - 27x^3 + O(x^5)$
$= (1+8-9)x + (\frac{1}{3} + \frac{32}{3} - 27)x^3 + O(x^5)$
$= 0x + (\frac{33}{3} - 27)x^3 + O(x^5) = (11 - 27)x^3 = -16x^3 + O(x^5)$
The denominator is $x^2 \tan x \approx x^2(x) = x^3$.
Thus,the limit is $\lim _{x \rightarrow 0} \frac{-16x^3}{x^3} = -16$.

(A) Given the limit $\lim _{x \rightarrow 0^{-}} \frac{\sin ^{-1}(x+[x])}{2-\{x\}}=\theta$.
For $x \rightarrow 0^{-}$,we have $[x] = -1$ and $\{x\} = x - [x] = x - (-1) = x+1$.
Substituting these values into the expression:
$\theta = \lim _{x \rightarrow 0^{-}} \frac{\sin ^{-1}(x-1)}{2-(x+1)} = \lim _{x \rightarrow 0^{-}} \frac{\sin ^{-1}(x-1)}{1-x}$.
As $x \rightarrow 0^{-}$,the expression approaches $\frac{\sin ^{-1}(-1)}{1-0} = \frac{-\pi/2}{1} = -\pi/2$.
Thus,$\theta = -\pi/2$.
Now,we calculate $\sin \theta + \cos \theta = \sin(-\pi/2) + \cos(-\pi/2) = -1 + 0 = -1$.

(B) Given $0 \leq a \leq 2$. The integral values of $a$ are $0, 1, 2$.
Let $f(x) = [x^2] - [x]^2$.
For $a = 0$:
$\text{L.H.L.} = \lim_{h \rightarrow 0} ([(0-h)^2] - [0-h]^2) = [h^2] - [-h]^2 = 0 - (-1)^2 = -1$.
$\text{R.H.L.} = \lim_{h \rightarrow 0} ([(0+h)^2] - [0+h]^2) = [h^2] - [h]^2 = 0 - 0 = 0$.
Since $\text{L.H.L.} \neq \text{R.H.L.}$,the limit does not exist at $a = 0$.
For $a = 1$:
$\text{L.H.L.} = \lim_{h \rightarrow 0} ([(1-h)^2] - [1-h]^2) = [1-2h+h^2] - [1-h]^2 = 0 - 0^2 = 0$.
$\text{R.H.L.} = \lim_{h \rightarrow 0} ([(1+h)^2] - [1+h]^2) = [1+2h+h^2] - [1+h]^2 = 1 - 1^2 = 0$.
Since $\text{L.H.L.} = \text{R.H.L.}$,the limit exists at $a = 1$.
For $a = 2$:
$\text{L.H.L.} = \lim_{h \rightarrow 0} ([(2-h)^2] - [2-h]^2) = [4-4h+h^2] - [2-h]^2 = 3 - 1^2 = 2$.
$\text{R.H.L.} = \lim_{h \rightarrow 0} ([(2+h)^2] - [2+h]^2) = [4+4h+h^2] - [2+h]^2 = 4 - 2^2 = 0$.
Since $\text{L.H.L.} \neq \text{R.H.L.}$,the limit does not exist at $a = 2$.
Thus,the limit does not exist for $2$ values of $a$ ($a=0$ and $a=2$).

(B) Given $f(x) = \frac{1-x+\sqrt{9x^2+10x+1}}{2x}$.
To find $\lim_{x \rightarrow -1^{-}} f(x)$,let $x = -1-h$,where $h \rightarrow 0^{+}$ as $x \rightarrow -1^{-}$.
Substituting $x = -1-h$ into the expression:
$f(-1-h) = \frac{1-(-1-h) + \sqrt{9(-1-h)^2 + 10(-1-h) + 1}}{2(-1-h)}$
$= \frac{2+h + \sqrt{9(1+2h+h^2) - 10 - 10h + 1}}{-2(1+h)}$
$= \frac{2+h + \sqrt{9+18h+9h^2 - 10 - 10h + 1}}{-2(1+h)}$
$= \frac{2+h + \sqrt{9h^2 + 8h}}{-2(1+h)}$
Taking the limit as $h \rightarrow 0^{+}$:
$\lim_{h \rightarrow 0^{+}} \frac{2+h + \sqrt{9h^2 + 8h}}{-2(1+h)} = \frac{2+0 + \sqrt{0}}{-2(1+0)} = \frac{2}{-2} = -1$.

(B) Assertion $(A)$: We know that $[x] = x - \{x\}$,where $\{x\}$ is the fractional part of $x$ and $0 \leq \{x\} < 1$.
$\lim_{x \rightarrow \infty} \frac{[x]}{x} = \lim_{x \rightarrow \infty} \frac{x - \{x\}}{x} = \lim_{x \rightarrow \infty} (1 - \frac{\{x\}}{x})$.
Since $0 \leq \{x\} < 1$,$\lim_{x \rightarrow \infty} \frac{\{x\}}{x} = 0$.
Thus,$\lim_{x \rightarrow \infty} \frac{[x]}{x} = 1 - 0 = 1$. So,$A$ is true.
Reason $(R)$: Given $f(x) = x - 1$ and $h(x) = x$.
$\lim_{x \rightarrow \infty} \frac{f(x)}{x} = \lim_{x \rightarrow \infty} \frac{x - 1}{x} = \lim_{x \rightarrow \infty} (1 - \frac{1}{x}) = 1$.
$\lim_{x \rightarrow \infty} \frac{h(x)}{x} = \lim_{x \rightarrow \infty} \frac{x}{x} = 1$.
Both limits are $1$,so $R$ is true. However,$R$ does not explain why $\lim_{x \rightarrow \infty} \frac{[x]}{x} = 1$.

(A) We are given the limit: $\lim _{x \rightarrow \frac{-3}{5}} \frac{1}{x}\left[\frac{-1}{x}\right]$.
As $x \rightarrow \frac{-3}{5}$,the term $\frac{-1}{x} \rightarrow \frac{-1}{-3/5} = \frac{5}{3}$.
Since $\frac{5}{3} = 1.66...$,it lies in the interval $[1, 2)$.
By the definition of the greatest integer function,$[\frac{5}{3}] = 1$.
Substituting these values into the expression,we get:
$\lim _{x}$ ${\rightarrow \frac{-3}{5}} \frac{1}{x}\left[\frac{-1}{x}\right] = \left(\frac{1}{-3/5}\right) \times [\frac{5}{3}] = \frac{-5}{3} \times 1 = \frac{-5}{3}$.

(C) Given the limit: $\lim _{x \rightarrow \infty} \frac{(b-c) x^2+(c-a) x+(a-b)}{(a-b) x^2+(b-c) x+(c-a)}=\frac{1}{2}$
Dividing the numerator and denominator by $x^2$:
$\lim _{x}$ ${\rightarrow \infty} \frac{(b-c) + \frac{c-a}{x} + \frac{a-b}{x^2}}{(a-b) + \frac{b-c}{x} + \frac{c-a}{x^2}} = \frac{1}{2}$
As $x \rightarrow \infty$,the terms with $x$ in the denominator approach $0$:
$\frac{b-c}{a-b} = \frac{1}{2}$
Cross-multiplying gives:
$2(b-c) = a-b$
$2b - 2c = a - b$
$3b = a + 2c$

(B) For $x \rightarrow a^{-}$,we have $\frac{x}{a} < 1$,so $\left[\frac{x}{a}\right] = 0$.
Thus,$k = \lim _{x \rightarrow a^{-}} \left(\frac{|x|^3}{a} - 0^3\right) = \frac{a^3}{a} = a^2$.
For $x \rightarrow a^{+}$,we have $\frac{x}{a} > 1$,so $\left[\frac{x}{a}\right] = 1$.
Thus,$l = \lim _{x \rightarrow a^{+}} \left(\frac{|x|^3}{a} - 1^3\right) = \frac{a^3}{a} - 1 = a^2 - 1$.
Calculating $k - l$:
$k - l = a^2 - (a^2 - 1) = a^2 - a^2 + 1 = 1$.

(B) Let $\lim _{n \rightarrow \infty} x_n = L$.
Since the limit exists and is finite,we can write the recurrence relation as $L = \frac{a + bL}{b + cL}$.
Multiplying both sides by $(b + cL)$,we get $L(b + cL) = a + bL$.
$bL + cL^2 = a + bL$.
Subtracting $bL$ from both sides,we get $cL^2 = a$.
$L^2 = \frac{a}{c}$.
Since $a, c > 0$ and the sequence terms are positive,we take the positive root: $L = \sqrt{\frac{a}{c}}$.

(C) We use the standard limit formula: $\lim _{x \rightarrow a} \frac{x^m - a^m}{x^n - a^n} = \frac{m}{n} a^{m-n}$.
Here,$m = \frac{1}{3}$,$n = \frac{1}{6}$,and $a = 1$.
Substituting these values into the formula:
$\lim _{z \rightarrow 1} \frac{z^{1/3} - 1^{1/3}}{z^{1/6} - 1^{1/6}} = \frac{1/3}{1/6} \times (1)^{1/3 - 1/6}$.
$= \frac{1}{3} \times \frac{6}{1} \times 1^{1/6}$.
$= 2 \times 1 = 2$.