Limits, Continuity and Differentiability

$ \begin{aligned} &\text { Using rationalisation }\\ &\begin{aligned} & \frac{(2 \tan x+\cos x-1+x)}{\left(\sqrt{4 \sin ^2 x+2 \tan x+1}\right.} \\ = & \lim _{x \rightarrow 0} \frac{\left.+\sqrt{3 \tan ^2 x+\sin x+1}\right)}{4 \sin ^2 x+2 \tan x-3 \tan ^2 x-\sin x} \\ = & \lim _{x \rightarrow 0} \frac{x\left[\frac{2 \tan x}{x}-\left(\frac{1-\cos x}{x^2}\right) x+1\right][\sqrt{1}+\sqrt{1}]}{\sin x(4 \sin x+2 \sec x-3 \sec x \tan x-1)} \\ = & \frac{(2-0+1)(2)}{(0+2-0-1)}=6 \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { } \lim _{x \rightarrow \frac{\pi}{4}} \frac{\cot ^3 x-\tan x}{\cos \left(x+\frac{\pi}{4}\right)} \\ & x=\frac{\pi}{4}+h \\ & =\lim _{h \rightarrow 0} \frac{1-\tan ^4\left(\frac{\pi}{4}+h\right)}{\tan ^3\left(\frac{\pi}{4}+h\right)(-\sin h)} \\ & =\lim _{h \rightarrow 0} \frac{\left(1+\tan ^2\left(\frac{\pi}{4}+h\right)\right.}{\tan ^3\left(\frac{\pi}{4}+h\right)} \frac{\left(1-\tan ^2\left(\frac{\pi}{4}+h\right)\right)}{-\sin h} \\ & =\lim _{h \rightarrow 0} \frac{1-\tan \left(\frac{\pi}{4}+h\right)}{-\sin h} \cdot\left[1+\tan \left(\frac{\pi}{4}+h\right)\right] \times 2 \end{aligned} $

$ \begin{aligned} & =\lim _{h \rightarrow 0} \frac{1-\tan \left(\frac{\pi}{4}+h\right)}{\left[1+\tan \left(\frac{\pi}{4}+h\right)\right]} \frac{\left[1+\tan \left(\frac{\pi}{4}+h\right)\right]^2}{[-\sin h]} \times 2 \\ & =\lim _{h \rightarrow 0} \frac{\tan \left(\frac{\pi}{4}-\frac{\pi}{4}-h\right)}{-\sin h} \times\left(2^2 \times 2\right. \\ & =\lim _{h \rightarrow 0} \frac{\tan (h)}{h} \frac{h}{\sin h} \times 8=8 \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { Let } L=\lim _{n \rightarrow \infty} \frac{1}{n^3} \sum_{k=1}^n k^2 x \\ & =\lim _{n \rightarrow \infty} \frac{1}{n^3}\left[1^2 x+2^2 x+3^2 x+\ldots+n^2 x\right] \\ & =\lim _{n \rightarrow \infty} \frac{1}{n^3} \times\left[1^2+2^2+3^2+\ldots+n^2\right] \\ & =\lim _{n \rightarrow \infty} \frac{x}{n^3} \cdot x \cdot \frac{n(n+1)(2 n+1)}{6} \\ & =\lim _{n \rightarrow \infty} \frac{1}{n^3} \times x \times \frac{n^3\left[\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)\right]}{6} \\ & =\frac{x}{6} \times 2=\frac{x}{3} \end{aligned} \end{aligned} $

The quadratic equation whose roots are $l, m$, where

$ \begin{aligned} l & =\lim _{\theta \rightarrow 0} \frac{3 \sin \theta-4 \sin ^3 \theta}{\theta} \\ & =\lim _{\theta \rightarrow 0} \frac{3 \sin \theta}{\theta}-4 \frac{\sin ^3 \theta}{\theta}=3 \\ l & =3 \\ m & =\lim _{\theta \rightarrow 0} 2 \frac{\tan \theta}{\theta\left(1-\tan ^2 \theta\right)} \\ & =\lim _{\theta \rightarrow 0} 2 \frac{\tan \theta}{\theta} \cdot \frac{1}{\left(1-\tan ^2 \theta\right)}=2 \\ m & =2 \end{aligned} $

Thus, the equation is $x^2-5 x+6=0$

Given, $\mathop {\lim }\limits_{x \to 2} \frac{\sqrt[3]{6+x}-\sqrt[3]{10-x}}{x-2}\left[\frac{0}{0}\right.$ form $]$ On applying L'Hospital's rule, we get

$ \begin{aligned} & =\lim _{x \rightarrow 2} \frac{\frac{1}{3}(6+x)^{-2 / 3}+\frac{1}{3}(10-x)^{-2 / 3}}{1} \\ & =\frac{1}{3}\left[(8)^{-2 / 3}+(8)^{-2 / 3}\right]=\frac{1}{3}\left[2(8)^{-2 / 3}\right] \\ & =\frac{1}{3}\left[2\left(2^3\right)^{-2 / 3}\right]=\frac{1}{3} \times 2 \times(2)^{-2} \\ & =\frac{1}{3} \times 2 \times \frac{1}{4}=\frac{1}{6} \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { Given, } \lim _{x \rightarrow 0} \frac{\tan ^4 x-\sin ^4 x}{x^6} \\ & =\lim _{x \rightarrow 0} \frac{\frac{\sin ^4 x}{\cos ^4 x}-\sin ^4 x}{x^2 \cdot x^4} \\ & =\lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)^4 \cdot \lim _{x \rightarrow 0} \frac{1-\cos ^4 x}{x^2 \cdot \cos ^4 x} \\ & =1 \cdot \lim _{x \rightarrow 0} \frac{\left(1-\cos ^2 x\right)\left(1+\cos ^2 x\right)}{x^2 \cos ^4 x} \\ & =\lim _{x \rightarrow 0} \frac{\sin ^2 x}{x^2} \frac{\left(1+\cos ^2 x\right)}{\cos ^4 x} \\ & =\lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)^2 \lim _{x \rightarrow 0} \frac{1+\cos ^2 x}{\cos ^4 x} \\ & =1 \times \frac{1+1}{1}=2 \end{aligned} \end{aligned} $

$ \text { } \begin{aligned} & \lim _{x \rightarrow-2^{+}}\left([x]^2-[x]-2\right)+\lim _{x \rightarrow-3^{-}}\left([x]^2-4[x]+3\right) \\ &=\left((-2)^2-(-2)-2\right)+\left((-4)^2-4(-4)+3\right) \\ &=(4+2-2)+(16+16+3)=39 \end{aligned} $

$ \begin{aligned} &\text { } \lim _{x \rightarrow 0} \frac{\left(3^{2 x}-\sqrt{x+1}\right) \sin 5 x}{1-\cos 4 x}\\ &=\lim _{x \rightarrow 0} \frac{3^{2 x}-\sqrt{x+1}}{x} \times \frac{\sin 5 x}{5 x} \times \frac{5 \cdot(2 x)^2}{2 \sin ^2 2 x} \times \frac{1}{4} \end{aligned} $

$ =\frac{5}{8}\left[\begin{array}{r} \lim _{x \rightarrow 0}\left(\frac{3^{2 x}-1}{x}+\frac{1-\sqrt{x+1}}{x}\right) \lim _{x \rightarrow 0} \frac{\sin 5 x}{5 x} \\ \lim _{x \rightarrow 0}\left(\frac{2 x}{\sin 2 x}\right)^2 \end{array}\right] $

$ \begin{aligned} & =\frac{5}{8}\left[2 \log _e 3-\frac{1}{2}\right]=\frac{5}{16}\left(\log _e 3^4-1\right) \\ & =\frac{5}{16} \log _e\left(\frac{81}{e}\right) \end{aligned} $

$\lim\limits_{x \rightarrow 1}(1-x) \tan \left(\frac{\pi}{2} x\right)$

Let $1-x=t \Rightarrow x=1-t$

So, as $\Rightarrow t \rightarrow 0$

$ \begin{aligned} &Now,\,\,\, \lim _{t \rightarrow 0} t \tan \left(\frac{\pi}{2}(1-t)\right) \\ & =\lim _{t \rightarrow 0} t \tan \left(\frac{\pi}{2}-\frac{\pi t}{2}\right)=\lim _{t \rightarrow 0} t \cot \left(\frac{\pi t}{2}\right) \\ & =\lim _{t \rightarrow 0} \frac{t}{\tan \left(\frac{\pi}{2} t\right)}=\frac{2}{\pi} \lim _{t \rightarrow 0}\left(\frac{\frac{\pi}{2} t}{\tan \frac{\pi t}{2}}\right) \\ & =\frac{2}{\pi} \cdot 1=\frac{2}{\pi} \end{aligned} $

Given, $f(9)=9$ and $f^{\prime}(9)=4$ and we have to find

$ \lim _{x \rightarrow 9} \frac{\sqrt{f(x)}-3}{\sqrt{x}-3} $

$ \begin{aligned} &\text { Apply } L^{\prime} \text { Hospital Rule, we get }\\ &\begin{aligned} \lim _{x \rightarrow 9} \frac{\frac{1}{2 \sqrt{f(x)}} \cdot f^{\prime}(x)}{\frac{1}{2 \sqrt{x}}} & =\frac{\frac{1}{\sqrt{f(9)}} f^{\prime}(9)}{\frac{1}{\sqrt{9}}} \\ & =\frac{4 / 3}{1 / 3}=4 \end{aligned} \end{aligned} $

$ \begin{aligned} & \lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{2 x^2+x-3} \\ & =\lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)(\sqrt{x}+1)}{\left(2 x^2+x-3\right)(\sqrt{x}+1)} \\ & =\lim _{x \rightarrow 1} \frac{(2 x-3)(x-1)}{(2 x+3)(x-1)(\sqrt{x}+1)} \\ & =\frac{(2-3)}{(2+3)} \times \frac{1}{(\sqrt{1}+1)}=-\frac{1}{5} \times \frac{1}{2}=\frac{-1}{10} \end{aligned} $

To solve the given problem, we need to evaluate the limit:

$ \lim_{x \rightarrow \infty} x\left(a^{1/x} + b^{1/x} + c^{1/x} - 3k^{1/x}\right) = 0 $

We start by substituting $x = \frac{1}{y}$, which implies $y \rightarrow 0$ as $x \rightarrow \infty$. The expression becomes:

$ \lim_{y \rightarrow 0} \frac{a^y + b^y + c^y - 3k^y}{y} = 0 $

Using the known limit:

$ \lim_{y \rightarrow 0} \frac{a^y - 1}{y} = \log_e a $

This allows us to rewrite the expression 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 $

This simplifies to:

$ \log_e a + \log_e b + \log_e c - 3 \log_e k = 0 $

Rewriting it gives:

$ \log_e \left(\frac{abc}{k^3}\right) = 0 $

Thus, we have:

$ \frac{abc}{k^3} = 1 $

Solving for $k$, we find:

$ k = (abc)^{1/3} $

$f(x)$ is continuous at $x = 0$

$\therefore$ $f(0) = \mathop {\lim }\limits_{x \to 0} f(x)$

$ \Rightarrow 10 = \mathop {\lim }\limits_{x \to 0} {{{{\log }_e}(1 + 5x) - {{\log }_e}(1 + \alpha x)} \over x}$

$ = \mathop {\lim }\limits_{x \to 0} {{\log (1 + 5x)} \over {5x}} \times 5 - {{{{\log }_e}(1 + \alpha x)} \over {\alpha x}} \times \alpha $

$ = 1 \times 5 - \alpha $

$ \Rightarrow \alpha = 5 - 10 = - 5$

$\mathop {\lim }\limits_{x \to 0} {{\alpha {e^x} + \beta {e^{ - x}} + \gamma \sin x} \over {x{{\sin }^2}x}} = {2 \over 3}$

$ \Rightarrow \alpha + \beta = 0$ (to make indeterminant form) ...... (i)

Now,

$\mathop {\lim }\limits_{x \to 0} {{\alpha {e^x} - \beta {e^{ - x}} + \gamma \cos x} \over {3{x^2}}} = {2 \over 3}$ (Using L-H Rule)

$ \Rightarrow \alpha - \beta + \gamma = 0$ (to make indeterminant form) ...... (ii)

Now,

$\mathop {\lim }\limits_{x \to 0} {{\alpha {e^x} + \beta {e^{ - x}} - \gamma \sin x} \over {6x}} = {2 \over 3}$ (Using L-H Rule)

$ \Rightarrow {{\alpha - \beta + \gamma } \over 6} = {2 \over 3}$

$ \Rightarrow \alpha - \beta + \gamma = 4$ ...... (iii)

$ \Rightarrow \gamma = - 2$

and (i) + (ii)

$2\alpha = - \gamma $

$ \Rightarrow \alpha = 1$ and $\beta = - 1$

and $\alpha {\beta ^2} + \beta {\gamma ^2} + \gamma {\alpha ^2} + 3 = 1 - 4 - 2 + 3 = - 2$

$f:R \to R$.

$f(x) = |x - 1|\cos |x - 2|\sin |x - 1| + (x - 3)|{x^2} - 5x + 4|$

$ = |x - 1|\cos |x - 2|\sin |x - 1| + (x - 3)|x - 1||x - 4|$

$ = |x - 1|[\cos |x - 2|\sin |x - 1| + (x - 3)|x - 4|]$

Sharp edges at $x = 1$ and $x = 4$

$\therefore$ Non-differentiable at $x = 1$ and $x = 4$

$f(x) = \mathop {\lim }\limits_{n \to \infty } {{\cos (2\pi x) - {x^{2n}}\sin (x - 1)} \over {1 + {x^{2n + 1}} - {x^{2n}}}}$

For $|x| < 1,\,f(x) = \cos 2\pi x$, continuous function

$|x| > 1,\,f(x) = \mathop {\lim }\limits_{n \to \infty } {{{1 \over {{x^{2n}}}}\cos 2\pi x - \sin (x - 1)} \over {{1 \over {{x^{2n}}}} + x - 1}}$

$ = {{ - \sin (x - 1)} \over {x - 1}}$, continuous

For $|x| = 1,\,f(x) = \left\{ {\matrix{ 1 & {\mathrm{if}} & {x = 1} \cr { - (1 + \sin 2)} & {\mathrm{if}} & {x = - 1} \cr } } \right.$

Now,

$\mathop {\lim }\limits_{x \to {1^ + }} f(x) = - 1,\,\mathop {\lim }\limits_{x \to {1^ - }} f(x) = 1$, so discontinuous at $x = 1$

$\mathop {\lim }\limits_{x \to {1^ + }} f(x) = 1,\,\mathop {\lim }\limits_{x \to - {1^ - }} f(x) = - {{\sin 2} \over 2}$, so discontinuous at $x = - 1$

$\therefore$ $f(x)$ is continuous for all $x \in R - \{ - 1,1\} $

$f(x) = {{\root 7 \of {p(729 + x)} - 3} \over {\root 3 \of {729 + qx} - 9}}$

for continuity at $x = 0$, $\mathop {\lim }\limits_{x \to 0} f(x) = f(0)$

Now, $\therefore$ $\mathop {\lim }\limits_{x \to 0} f(x) = \mathop {\lim }\limits_{x \to 0} {{\root 7 \of {p(729 + x)} - 3} \over {\root 3 \of {729 + qx} - 9}}$

$ \Rightarrow p = 3$ (To make indeterminant form)

So, $\mathop {\lim }\limits_{x \to 0} f(x) = \mathop {\lim }\limits_{x \to 0} {{{{\left( {{3^7} + 3x} \right)}^{{1 \over 7}}} - 3} \over {{{\left( {729 + qx} \right)}^{{1 \over 3}}} - 9}}$

$ = \mathop {\lim }\limits_{x \to 0} {{3\left[ {{{\left( {1 + {x \over {{3^6}}}} \right)}^{{1 \over 7}}} - 1} \right]} \over {9\left[ {{{\left( {1 + {q \over {729}}x} \right)}^{{1 \over 3}}} - 1} \right]}} = {1 \over 3}\,.\,{{{1 \over 7}\,.\,{1 \over {{3^6}}}} \over {{1 \over 3}\,.\,{q \over {729}}}}$

$\therefore$ $f(0) = {1 \over {7q}}$

$\therefore$ Option (B) is correct.

$\beta = \mathop {\lim }\limits_{x \to 0} {{\alpha x - ({e^{3x}} - 1)} \over {\alpha x({e^{3x}} - 1)}},\,\alpha \in R$

$ = \mathop {\lim }\limits_{x \to 0} {{{\alpha \over 3} - \left( {{{{e^{3x}} - 1} \over {3x}}} \right)} \over {\alpha x\left( {{{{e^{3x - 1}}} \over {3x}}} \right)}}$

So, $\alpha = 3$ (to make independent form)

$\beta = \mathop {\lim }\limits_{x \to 0} {{1 - \left( {{{{e^{3x}} - 1} \over {3x}}} \right)} \over {3x}} = {{1 - {{\left( {3x + {{9{x^2}} \over 2}\, + \,......} \right)} \over {3x}}} \over {3x}}$

$ = {{ - \left( {{9 \over 2}{x^2} + {{{{(3x)}^3}} \over {31}}\, + \,....} \right)} \over {9{x^2}}} = {{ - 1} \over 2}$

$\therefore$ $\alpha + \beta = 3 - {1 \over 2} = {5 \over 2}$

$f(3x) - f(x) = x$ ...... (1)

$x \to {x \over 3}$

$f(x) - f\left( {{x \over 3}} \right) = {x \over 3}$ ....... (2)

Again $x \to {x \over 3}$

$f\left( {{x \over 3}} \right) - f\left( {{x \over 9}} \right) = {x \over {{3^2}}}$ ...... (3)

Similarly

$f\left( {{x \over {{3^{n - 2}}}}} \right) - f\left( {{x \over {{3^{n - 1}}}}} \right) = {x \over {{3^{n - 1}}}}\,.....\,(n)$

Adding all these and applying $n \to \infty $

$\mathop {\lim }\limits_{n \to \infty } \left( {f(3x) - f\left( {{x \over {{3^{n - 1}}}}} \right)} \right) = x\left( {1 + {1 \over 3} + {1 \over {{3^2}}}\, + \,....} \right)$

$f(3x) - f(0) = {{3x} \over 2}$

Putting $x = {8 \over 3}$

$f(8) - f(0) = 4$

$ \Rightarrow f(0) = 3$

Putting $x = {{14} \over 3}$

$f(14) - 3 = 7 \Rightarrow f(14) = 10$

$f(x) = \left\{ {\matrix{ {{{{{\log }_e}(1 - x + {x^2}) + {{\log }_e}(1 + x + {x^2})} \over {\sec x - \cos x}}} & , & {x \in \left( {{{ - \pi } \over 2},{\pi \over 2}} \right) - \{ 0\} } \cr k & , & {x = 0} \cr } } \right.$

for continuity at $x = 0$

$\mathop {\lim }\limits_{x \to 0} f(x) = k$

$\therefore$ $k = \mathop {\lim }\limits_{x \to 0} {{{{\log }_e}({x^4} + {x^2} + 1)} \over {\sec x - \cos x}}\left( {{0 \over 0}\,\mathrm{form}} \right)$

$ = \mathop {\lim }\limits_{x \to 0} {{\cos x{{\log }_e}({x^4} + {x^2} + 1)} \over {{{\sin }^2}x}}$

$ = \mathop {\lim }\limits_{x \to 0} {{{{\log }_e}({x^4} + {x^2} + 1)} \over {{x^2}}}$

$ = \mathop {\lim }\limits_{x \to 0} {{\ln (1 + {x^2} + {x^4})} \over {{x^2} + {x^4}}}\,.\,{{{x^2} + {x^4}} \over {{x^2}}}$

$ = 1$

$f(x) = \left\{ {\matrix{ {x + a} & , & {x \le 0} \cr {|x - 4|} & , & {x > 0} \cr } } \right.$ and $g(x) = \left\{ {\matrix{ {x + 1} & , & {x < 0} \cr {{{(x - 4)}^2} + b} & , & {x \ge 0} \cr } } \right.$

$\because$ $f(x)$ and $g(x)$ are continuous on R

$\therefore$ $a = 4$ and $b = 1 - 16 = - 15$

then $(gof)(2) + (fog)( - 2)$

$ = g(2) + f( - 1)$

$ = - 11 + 3 = - 8$

$f(x) = \left\{ {\matrix{ {{x^3} - {x^2} + 10x - 7,} & {x \le 1} \cr { - 2x + {{\log }_2}({b^2} - 4),} & {x > 1} \cr } } \right.$

If $f(x)$ has maximum value at $x = 1$ then $f(1 + ) \le f(1)$

$ - 2 + {\log _2}({b^2} - 4) \le 1 - 1 + 10 - 7$

${\log _2}({b^2} - 4) \le 5$

$0 < {b^2} - 4 \le 32$

(i) ${b^2} - 4 > 0 \Rightarrow b \in ( - \infty , - 2) \cup (2,\infty )$

(ii) ${b^2} - 36 \le 0 \Rightarrow b \in [ - 6,6]$

Intersection of above two sets

$b \in [ - 6, - 2) \cup (2,6]$

$\mathop {\lim }\limits_{x \to {\pi \over 4}} {{8\sqrt 2 - {{(\cos x + \sin x)}^7}} \over {\sqrt 2 - \sqrt 2 \sin 2x}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{0 \over 0}\,\mathrm{form}} \right)$

$ = \mathop {\lim }\limits_{x \to {\pi \over 4}} {{ - 7{{(\cos x + \sin x)}^6}( - \sin x + \cos x)} \over { - 2\sqrt 2 \cos 2x}}$ using $\mathrm{L-H}$ Rule

$ = \mathop {\lim }\limits_{x \to {\pi \over 4}} {{56(\cos x - \sin x)} \over {2\sqrt 2 \cos 2x}}\,\,\left( {{0 \over 0}} \right)$

$ = \mathop {\lim }\limits_{x \to {\pi \over 4}} {{ - 56(\sin x + \cos x)} \over { - 4\sqrt 2 \sin 2x}}$ using $\mathrm{L-H}$ Rule

$ = 7\sqrt 2 \,.\,\sqrt 2 = 14$

$\mathop {\lim }\limits_{n \to \alpha } \left( {\sqrt {{n^2} - n - 1} + n\alpha + \beta } \right) = 0$

[ This limit will be zero when $\alpha$ < 0 as when $\alpha$ > 0 then overall limit will be $\infty $. ]

$ \Rightarrow \mathop {\lim }\limits_{n \to \alpha } {{\left( {\sqrt {{n^2} - n - 1} + n\alpha + \beta } \right)\left( {\sqrt {{n^2} - n - 1} - \left( {n\alpha + \beta } \right)} \right)} \over {\sqrt {{n^2} - n - 1} - \left( {n\alpha + \beta } \right)}} = 0$

$ \Rightarrow \mathop {\lim }\limits_{n \to \alpha } {{\left( {{n^2} - n - 1} \right) - {{\left( {n\alpha + \beta } \right)}^2}} \over {\sqrt {{n^2} - n - 1} - \left( {n\alpha + \beta } \right)}} = 0$

$ \Rightarrow \mathop {\lim }\limits_{n \to \alpha } {{{n^2} - n - 1 - {n^2}{\alpha ^2} - 2n\alpha\beta - {\beta ^2}} \over {\sqrt {{n^2} - n - 1} - \left( {n\alpha + \beta } \right)}} = 0$

$ \Rightarrow \mathop {\lim }\limits_{n \to \alpha } {{{n^2}\left( {1 - {\alpha ^2}} \right) - n\left( {1 + 2\alpha \beta } \right) - \left( {1 + {\beta ^2}} \right)} \over {\sqrt {{n^2} - n - 1} - \left( {n\alpha + \beta } \right)}}$

Here power of "n" in the numerator is 2 and power of "n" in the denominator is 1.

To get the value of limit equal to zero power of "n" should be equal in both numerator and denominator, otherwise value of limit will be infinite ($\infty $).

$\therefore$ Coefficient of n² should be 0 in this case.

$\therefore$ $1 - {\alpha ^2} = 0$

$ \Rightarrow \alpha = \, \pm \,1$

But $\alpha$ should be < 0

$\therefore$ $\alpha = \, + \,1$ not possible

$\therefore$ $\alpha = - 1$.

$ \Rightarrow \mathop {\lim }\limits_{n \to \alpha } {{0 - n\left( {1 + 2\alpha \beta } \right) - \left( {1 + \beta } \right)} \over {n\left[ {\sqrt {1 - {1 \over n} - {1 \over {{n^2}}}} - \alpha - {\beta \over n}} \right]}} = 0$

Divide numerator and denominator by n then we get,

$ \Rightarrow \mathop {\lim }\limits_{n \to \alpha } {{ - \left( {1 + 2\alpha \beta } \right) - {{(1 + \beta )} \over n}} \over {\sqrt {1 - {1 \over n} - {1 \over {{n^2}}}} - \alpha - {\beta \over n}}} = 0$

$ \Rightarrow {{ - \left( {1 + 2\alpha \beta } \right) - 0} \over {\sqrt {1 - 0 - 0} - \alpha - 0}} = 0$

$ \Rightarrow {{ - \left( {1 + 2\alpha \beta } \right)} \over {1 - \alpha }} = 0$

$ \Rightarrow - \left( {1 + 2\alpha \beta } \right) = 0$

$ \Rightarrow 1 + 2\alpha \beta = 0$

$ \Rightarrow 2\alpha \beta = - 1$

$ \Rightarrow \beta = - {1 \over {2\alpha }} = - {1 \over {2( - 1)}} = {1 \over 2}$

$\therefore$ $8\left( {\alpha + \beta } \right)$

$ = 8\left( { - 1 + {1 \over 2}} \right)$

$ = 8 \times - {1 \over 2}$

$ = - 4$

$\mathop {\lim }\limits_{x \to 1} {{({x^2} - 1){{\sin }^2}(\pi x)} \over {{x^4} - 2{x^2} + 2x - 1}}$

$ = \mathop {\lim }\limits_{x \to 1} {{({x^2} - 1)si{n^2}(\pi x)} \over {({x^2} - 1){{(x - 1)}^2}}}$

$ = \mathop {\lim }\limits_{x \to 1} {{{{\sin }^2}(\pi x)} \over {{{(x - 1)}^2}}}$

Let $x = 1 + h$

$\therefore$ when x $\to$ 1 then h $\to$ 0

$ = \mathop {\lim }\limits_{h \to 0} {{{{\sin }^2}(\pi (1 + h))} \over {{{(1 + h - 1)}^2}}}$

$ = \mathop {\lim }\limits_{h \to 0} {{{{\sin }^2}(\pi h)} \over {{h^2}}}$

$ = \mathop {\lim }\limits_{h \to 0} {\pi ^2} \times {{{{\sin }^2}(\pi h)} \over {{{(\pi h)}^2}}}$

$ = {\pi ^2} \times 1$

$ = {\pi ^2}$

$\mathop {\lim }\limits_{n \to \infty } 6\tan \left\{ {\sum\limits_{r = 1}^n {{{\tan }^{ - 1}}\left( {{1 \over {{r^2} + 3r + 3}}} \right)} } \right\}$

$ = \mathop {\lim }\limits_{n \to \infty } 6\tan \left\{ {\sum\limits_{r = 1}^n {{{\tan }^{ - 1}}\left( {{{(r + 2) - (r + 1)} \over {1 + (r + 2)(r + 1)}}} \right)} } \right\}$

$ = \mathop {\lim }\limits_{n \to \infty } 6\tan \left\{ {\sum\limits_{r = 1}^n {({{\tan }^{ - 1}}(r + 2) - {{\tan }^{ - 1}}(r + 1))} } \right\}$

$ = \mathop {\lim }\limits_{n \to \infty } 6\tan \left\{ {{{\tan }^{ - 1}}(n + 2) - {{\tan }^{ - 1}}2} \right\}$

$ = 6\tan \left\{ {{\pi \over 2} - {{\cot }^{ - 1}}\left( {{1 \over 2}} \right)} \right\}$

$ = 6\tan \left( {{{\tan }^{ - 1}}\left( {{1 \over 2}} \right)} \right)$

$ = 3$

$f(x) = \left\{ {\matrix{ 0 & {x < 0} \cr {a{e^x} - 1} & {0 \le x < 1} \cr b & {x = 1} \cr {b - 1} & {1 < x < 2} \cr { - c} & {x \ge 2} \cr } } \right.$

To be continuous at x = 0

a $-$ 1 = 0

to be continuous at x = 1

ae $-$ 1 = b = b $-$ 1 $\Rightarrow$ not possible

to be continuous at x = 2

b $-$ 1 = $-$ c $\Rightarrow$ b + c = 1

If a = 1 and b + c = 1 then f(x) is discontinuous at exactly one point.

$\mathop {\lim }\limits_{x \to 7} {{18 - [1 - x]} \over {[x - 3a]}}$ exist & $a \in I$.

$ = \mathop {\lim }\limits_{x \to 7} {{17 - [ - x]} \over {[x] - 3a}}$ exist

$RHL = \mathop {\lim }\limits_{x \to {7^ + }} {{17 - [ - x]} \over {[x] - 3a}} = {{25} \over {7 - 3a}}$ $\left[ {a \ne {7 \over 3}} \right]$

$LHL = \mathop {\lim }\limits_{x \to {7^ - }} {{17 - [ - x]} \over {[x] - 3a}} = {{24} \over {6 - 3a}}$ $\left[ {a \ne 2} \right]$

For limit to exist

$LHL = RHL$

${{25} \over {7 - 3a}} = {{24} \over {6 - 3a}}$

$ \Rightarrow {{25} \over {7 - 3a}} = {8 \over {2 - a}}$

$\therefore$ $a = - 6$

$\mathop {\lim }\limits_{x \to 0} {{\cos (\sin x) - \cos x} \over {{x^4}}} = \mathop {\lim }\limits_{x \to 0} {{2\sin (x + \sin x)\,.\,\sin \left( {{{x - \sin x} \over 2}} \right)} \over {{x^4}}}$

$ = \mathop {\lim }\limits_{x \to 0} 2\,.\,\left( {{{\left( {{{x + \sin x} \over 2}} \right)\left( {{{x - \sin x} \over 2}} \right)} \over {{x^4}}}} \right)$

$ = \mathop {\lim }\limits_{x \to 0} {1 \over 2}\,.\,\left( {{{\left( {x + x - {{{x^3}} \over {3!}} + {{{x^5}} \over {5!}}...} \right)\left( {x - x + {{{x^3}} \over {3!}}...} \right)} \over {{x^4}}}} \right)$

$ = \mathop {\lim }\limits_{x \to 0} {1 \over 2}\,.\,\left( {2 - {{{x^2}} \over {3!}} + {{{x^4}} \over {5!}}...} \right)\left( {{1 \over {3!}} - {{{x^2}} \over {5!}} - 1} \right)$

$ = {1 \over 6}$

$f(x) = \min \{ 1,\,1 + x\sin x\} $, $0 \le x \le x$

$f(x) = \left\{ {\matrix{ {1,} & {0 \le x < \pi } \cr {1 + x\sin x,} & {\pi \le x \le 2\pi } \cr } } \right.$

Now at $x = \pi ,\,\,\mathop {\lim }\limits_{x \to {\pi ^ - }} f(x) = 1 = \mathop {\lim }\limits_{x \to {\pi ^ + }} f(x)$

$\therefore$ f(x) is continuous in [0, 2$\pi$]

Now, at x = $\pi$ $L.H.D = \mathop {\lim }\limits_{h \to 0} {{f(\pi - h) - f(\pi )} \over { - h}} = 0$

$R.H.D = \mathop {\lim }\limits_{h \to 0} {{f(\pi + h) - f(\pi )} \over h} = 1 - {{(\pi + h)\sin \,h - 1} \over h} = - \pi $

$\therefore$ f(x) is not differentiable at x = $\pi$

$\therefore$ (m, n) = (1, 0)

$\mathop {\lim }\limits_{x \to {1 \over {\sqrt 2 }}} {{\sin ({{\cos }^{ - 1}}x) - x} \over {1 - \tan ({{\cos }^{ - 1}}x)}}$

Let ${\cos ^{ - 1}}x = t$

$ \Rightarrow x = \cos t$

When $x \to {1 \over {\sqrt 2 }}$, then $t \to {\cos ^{ - 1}}\left( {{1 \over {\sqrt 2 }}} \right) \to {\pi \over 4}$

$\therefore$ $\mathop {\lim }\limits_{t \to {\pi \over 4}} {{\sin t - \cos t} \over {1 - \tan (t)}}$

$ = \mathop {\lim }\limits_{t \to {\pi \over 4}} {{\sin t - \cos t} \over {1 - {{\sin t} \over {\cos t}}}}$

$ = \mathop {\lim }\limits_{t \to {\pi \over 4}} {{(\sin t - \cos t)(\cos t)} \over {(\cos t - \sin t)}}$

$ = \mathop {\lim }\limits_{t \to {\pi \over 4}} - \cos t$

$ = - \mathop {\lim }\limits_{t \to {\pi \over 4}} \cos t$

$ = - {1 \over {\sqrt 2 }}$

$\because$ gof is differentiable at x = 0

So R.H.D = L.H.D

${d \over {dx}}(4{e^x} + {k_2}) = {d \over {dx}}\left( {{{( - |x + 3|)}^2} - {k_1}|x + 3|} \right)$

$ \Rightarrow 4 = 6 - {k_1} \Rightarrow {k_1} = 2$

Also $f(f({0^ + })) = g(f({0^ - }))$

$ \Rightarrow 4 + {k_2} = 9 - 3{k_1} \Rightarrow {k_2} = - 1$

Now $g(f( - 4)) + g(f(4))$

$ = g( - 1) + g({e^4}) = (1 - {k_1}) + (4{e^4} + {k_2})$

$ = 4{e^4} - 2$

$ = 2(2{e^4} - 1)$

$\mathop {\lim }\limits_{x \to {\pi \over 2}} {\tan ^2}x\left\{ {\sqrt {2{{\sin }^2}x + 3\sin x + 4} - \sqrt {{{\sin }^2}x + 6\sin x + 2} } \right\}$

$ = \mathop {\lim }\limits_{x \to {\pi \over 2}} {{{{\tan }^2}x({{\sin }^2}x - 3\sin x + 2)} \over {\sqrt {2{{\sin }^2}x + 3\sin x + 4} + \sqrt {{{\sin }^2}x + 6\sin x + 2} }}$

$ = {1 \over 6}\mathop {\lim }\limits_{x \to {\pi \over 2}} {{(1 - \sin x)(2 - \sin x)} \over {{{\cos }^2}x}}\,.\,{\sin ^2}x$

$ = {1 \over 6}\mathop {\lim }\limits_{x \to {\pi \over 2}} {{(2 - \sin x){{\sin }^2}x} \over {1 + \sin x}}$

$ = {1 \over {12}}$

Given, $f(x) + f'(x) + f''(x) = {x^5} + 64$ .........(i)

$\Rightarrow f(x)$ is a polynomial in $x$ whose degree is 5.

Let $f(x) = {x^5} + a{x^4} + b{x^3} + c{x^2} + dx + e$

$f'(x) = 5{x^4} + 4a{x^3} + 3b{x^2} + 2cx + d$

$f''(x) = 20{x^3} + 12a{x^2} + 6bx + 2c$

On substituting the value of $f(x), f^{\prime}(x)$ and $f^{\prime \prime}(x)$ in Eq. (i), we get

${x^5}+(a + 5){x^4} + (b + 4a + 20){x^3} + (c + 3b + 12a){x^2} + (d + 2c + 6b)x + e + d + 2c = {x^5} + 64$

Now, equating the coefficient, we get

$ \Rightarrow a + 5 = 0$

$b + 4a + 20 = 0$

$c + 3b + 12a = 0$

$d + 2c + 6b = 0$

$e + d + 2c = 64$

$\therefore$ $a = - 5,\,b = 0,\,c = 60,\,d = - 120,\,e = 64$

$\therefore$ $f(x) = {x^5} - 5{x^4} + 60{x^2} - 120x + 64$

Now, $\mathop {\lim }\limits_{x \to 1} {{{x^5} - 5{x^4} + 60{x^2} - 120x + 64} \over {x - 1}}$ is (${0 \over 0}$ form)

By L' Hospital rule

$\mathop {\lim }\limits_{x \to 1} {{5{x^4} - 20{x^3} + 120x - 120} \over 1}$

$ = - 15$

$f(x) = \left\{ {\matrix{ {{{\sin (x - [x])} \over {x[x]}}} & , & {x \in ( - 2, - 1)} \cr {\max \{ 2x,3[|x|]\} } & , & {|x| < 1} \cr 1 & , & {otherwise} \cr } } \right.$

$f(x) = \left\{ {\matrix{ {{{\sin (x + 2)} \over {x + 2}}} & , & {x \in ( - 2, - 1)} \cr 0 & , & {x \in ( - 1,0]} \cr {2x} & , & {x \in (0,1)} \cr 1 & , & {otherwise} \cr } } \right.$

It clearly shows that f(x) is discontinuous

At x = $-$1, 1 also non differentiable

and at $x = 0$, $L.H.D = \mathop {\lim }\limits_{h \to 0} {{f(0 - h) - f(0)} \over { - h}} = 0$

$R.H.D = \mathop {\lim }\limits_{h \to 0} {{f(0 + h) - f(0)} \over h} = 2$

$\therefore$ f(x) is not differentiable at x = 0

$\therefore$ m = 2, n = 3

$ \text { } \mathop {\lim }\limits_{x \to 2} \frac{x^3-x^2-x-2}{2 x^3-3 x^2-3 x+2} $

$ \begin{aligned} & \therefore \lim _{x \rightarrow 2} \frac{x^3-x^2-x-2}{2 x^3-3 x^2-3 x+2} \\ & =\lim _{x \rightarrow 2} \frac{(x-2)\left(x^2+x+1\right)}{(x-2)\left(2 x^2+x-1\right)} \\ & =\lim _{x \rightarrow 2} \frac{x^2+x+1}{2 x^2+x-1}=\frac{4+2+1}{8+2-1}=\frac{7}{9} \end{aligned} $

$ \begin{aligned} & \text { } \lim _{x \rightarrow 0} \frac{4[\sin (2022 x)-\sin (2020 x)]}{x[\cos (2022 x)+2 \cos (2021 x)} +\cos (2020 x)] \\ & =\lim _{x \rightarrow 0} 4\left[\frac{\sin 2022 x}{x}-\frac{\sin 2020 x}{x}\right] \\ & \times\left[\frac{1}{\cos 2022 x+2 \cos 2021 x+\cos 2020 x}\right] \end{aligned} $

$ =\mathop {\lim }\limits_{x \to 0} 4\left[\begin{array}{l} 2022 \cdot\left(\frac{\sin 2022 x}{2022 x}\right) -2020\left(\frac{\sin 2020 x}{2020 x}\right) \end{array}\right] $

$ \begin{aligned} & \times\left[\frac{1}{\cos 2022 x+2 \cos 2021 x+\cos 2020 x}\right] \\ &= 4[2022 \times 1-2020 \times 1] \\ & \times\left[\frac{1}{\cos 0+2 \cos 0+\cos 0}\right] \\ &= 4(2022-2020) \times\left[\frac{1}{1+2+1}\right] \\ &= 4 \times 2 \times \frac{1}{4}=2 \end{aligned} $