Limits, Continuity and Differentiability

$\lim _\limits{x \rightarrow 2^{-}} f(x)=f(2)$

$\because f(x)$ is continuous on $[0,8]$.

$\Rightarrow f(x)$ is continuous at 2 and 4 as well

$ \begin{aligned} 4+2 a+b & =8 \\ \Rightarrow 2 a+b & =4 \quad .....\text{(i)}\\ \lim _{x \rightarrow 4^{+}} f(x) & =f(4) \\ 8 a+5 b & =14\quad .....\text{(ii)} \end{aligned}$

From Eqs. (i) and (ii), we get

$a=3, b=-2$

$\mathop {\lim }\limits_{x \to {4^ - }} f(x) = f(4) = \mathop {\lim }\limits_{x \to {4^ - }} f(x)$

$ - 1 + a = a + b = 1 + b$

$ \Rightarrow b = - 1,a = 1$

$f$ is continuous $\Rightarrow f(x)$ will be continuous at $x=0$

\begin{array}{ll} \Rightarrow & \lim _\limits{x \rightarrow 0} f(x)=f(0) \\ \Rightarrow & \lim _\limits{x \rightarrow 0} \frac{e^{\alpha x}-e^x-x}{x^2}=\frac{3}{2} \\ \Rightarrow & \lim _\limits{x \rightarrow 0} \frac{\alpha e^{\alpha x}-e^x-1}{2 x}=\frac{3}{2}\end{array}$

[By $L$ hospital's rule]

This is solvable when

$\alpha e^0-e^0-1=0 \Rightarrow \alpha-2=0$

or $\alpha=2$

$\begin{aligned} & \text { } f(0)=\lim _{x \rightarrow 0} \frac{\left(e^x-1\right)^4}{\sin \left(\frac{x^2}{k^2}\right) \log \left\{1+\frac{x^2}{2}\right\}}=8 \\ & \Rightarrow \lim _{x \rightarrow 0} \frac{2 k^2\left(\frac{e^x-1}{x}\right)^4}{\frac{\sin \left(\frac{x^2}{k^2}\right)}{\left(\frac{x^2}{k^2}\right)} \cdot \frac{\log \left\{1+\frac{x^2}{2}\right\}}{\frac{x^2}{2}}}=8 \\ & \end{aligned}$

$\begin{aligned} & \left\{\begin{array}{l}\because \lim _\limits{x \rightarrow} \frac{e^x-1}{x}=1 \Rightarrow \lim _\limits{x \rightarrow 0} \frac{\sin x}{x}=1 \\ \lim _\limits{x \rightarrow 0} \frac{\log (1+x)}{x}=1\end{array}\right\} \\ & \Rightarrow \quad 2 k^2=8 \\ & \Rightarrow \quad k= \pm 2 \\ & \end{aligned}$

$\begin{aligned} & \text { } \lim _{x \rightarrow 0} \frac{2 f(x)-3 f(2 x)+f(4 x)}{x^2} \quad \text{[0/0 form]}\\ & =\lim _{x \rightarrow 0} \frac{2 f^{\prime}(x)-6 f^{\prime}(2 x)+4 f^{\prime}(4 x)}{2 x} \quad \text{[0/0 form]}\\ & =\lim _{x \rightarrow 0} \frac{2 f^{\prime \prime}(x)-12 f^{\prime \prime}(2 x)+16 f^{\prime \prime}(4 x)}{2} \\ & =\frac{2 f^{\prime \prime}(0)-12 f^{\prime \prime}(0)+16 f^{\prime \prime}(0)}{2} \\ & =3 f^{\prime \prime}(0)=3 \cdot 4=12 \\ & {\left[\because f^{\prime \prime}(0)=4\right]} \\ \end{aligned}$

$\lim _\limits{z \rightarrow 1} \frac{z^{1 / 3}-1}{z^{1 / 6}-1}$ $\left[\frac{0}{0} \text { form }\right]$

$\begin{aligned} & =\lim _{z \rightarrow 1} \frac{\frac{1}{3} z^{-2 / 3}}{\frac{1}{6} z^{-5 / 6}} \text {[L' Hospital rule]}\\ & =\frac{6}{3} \frac{(1)^{5 / 6}}{(1)^{2 / 3}}=\frac{6}{3}=2 \end{aligned}$

$f$ is continuous at $x=0$

$\begin{aligned} & \therefore \lim_\limits{x \rightarrow 0} f(x)=f(0)\\ & \Rightarrow \lim _\limits{x \rightarrow 0} \frac{72^x-9^x-8^x+1}{\sqrt{2}-\sqrt{1+\cos x}}=k \log 2 \log 3 \\ & \Rightarrow k \log 2 \cdot \log 3 \\ & =\lim _{x \rightarrow 0} \frac{\left(9^x-1\right)\left(8^x-1\right)(\sqrt{2}+\sqrt{1+\cos x})}{2-(1+\cos x)} \end{aligned}$

$\begin{aligned} & =\lim _{x \rightarrow 0}\left[\begin{array}{c}\frac{\left(9^x-1\right)}{x} \cdot\left(\frac{8^x-1}{x}\right) \cdot \frac{x^2}{1-\cos x} \\ \cdot(\sqrt{2}+\sqrt{1+\cos x})\end{array}\right] \\ & =\log 9 \cdot \log 8 \cdot 2(\sqrt{2}+\sqrt{2}) \\ & =2 \log 3 \cdot 3 \log 2 \cdot 4 \sqrt{2} \\ & \Rightarrow k \log 3 \cdot \log 2=24 \sqrt{2} \log 3 \cdot \log 2 \Rightarrow k=24 \sqrt{2} \\ & \end{aligned}$

$\because f$ is continuous in $(0, \pi) . \Rightarrow f$ is continuous at $x=\frac{\pi}{4}, \frac{\pi}{2}$

$\therefore$ LHL of $f(x)\left(\right.$ at $\left.x=\frac{\pi}{4}\right)=$ RHL of $f(x)\left(\right.$ at $\left.x=\frac{\pi}{4}\right) =f\left(\frac{\pi}{4}\right)$

$\begin{array}{r} \Rightarrow \quad \lim _\limits{x \rightarrow \frac{\pi^{-}}{4}} f(x)=\lim _\limits{x \rightarrow \frac{\pi^{+}}{4}} f(x)=2\left(\frac{\pi}{4}\right) \cot \frac{\pi}{4}+b \\ \Rightarrow \quad \lim _\limits{x \rightarrow \frac{\pi^{-}}{4}}(x+a \sqrt{2} \sin x)=\lim _\limits{x \rightarrow \frac{\pi^{+}}{4}}(2 x \cot x+b) \\ =\frac{\pi}{2}+b\end{array}$

$\Rightarrow \frac{\pi}{4}+a \sqrt{2} \sin \frac{\pi}{4}=2 \cdot \frac{\pi}{4} \cot \frac{\pi}{4}+b$

$\Rightarrow a-b=\frac{\pi}{4}$ ..... (i)

Again, LHL of $f(x)\left(\right.$ at $\left.x=\frac{\pi}{2}\right)=$ RHL of $f(x)\left(\right.$ at $\left.x=\frac{\pi}{2}\right) \quad\left\{\because f(x)\right.$ is continuous at $\left.x=\frac{\pi}{2}\right\}$ $ \begin{array}{ll} \Rightarrow \quad & \lim _\limits{x \rightarrow \frac{\pi^{-}}{2}} f(x)=\lim _\limits{x \rightarrow \frac{\pi^{+}}{2}} f(x) \\ \Rightarrow \quad & \lim _\limits{x \rightarrow \frac{\pi^{-}}{2}}(2 x \cot x+b) \\ & =\lim _\limits{x \rightarrow \frac{\pi^{+}}{2}}(a \cos 2 x-b \sin x) \\ \Rightarrow \quad & 2 \cdot \frac{\pi}{2} \cot \frac{\pi}{2}+b=a \cos \pi-b \sin \frac{\pi}{2} \end{array}$

$\Rightarrow a=-2b$ ..... (ii)

On solving Eqs. (i) and (ii), we get

$a=\frac{\pi}{6},b=-\frac{\pi}{12}$

The required limit

$ = \mathop {\lim }\limits_{x \to 0} {{\sqrt {(1 - \cos {x^2})(1 + \cos {x^2})} } \over {(1 - \cos x)\sqrt {1 + \cos {x^2}} }}$

$ = \mathop {\lim }\limits_{x \to 0} {{\left( {{{\sin {x^2}} \over {{x^2}}}} \right)} \over {2\left( {{{{{\sin }^2}x/2} \over {{x^2}}}} \right)}}.{1 \over {\sqrt {1 + \cos {x^2}} }} = {1 \over {2 \times {1 \over 4} \times \sqrt 2 }} = \sqrt 2 $

$f'(1 - 0) = \mathop {\lim }\limits_{h \to 0} {{f(1 - h) - f(1)} \over { - h}}$

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

$f'(1 + 0) = \mathop {\lim }\limits_{h \to 0} {{f(1 + h) - f(1)} \over h}$

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

$ = \mathop {\lim }\limits_{h \to 0} {{{h^2} + 2h + ah + b} \over h} = 2 + a$, if b = 0

Thus, $2a = 2 + a,b = 0$

$ \Rightarrow a = 2,b = 0$

From graph you can see,
(1) when x < 0 then y = x² is greater than y = x. That is why for f(x) that curved part is chosen.
(2) when 0 $ \le $ x < 1 then y = x is greater than y = x². That is why for f(x) part of that straight line is chosen.
(3) when x $ \ge $ 1 then y = x² is greater than y = x. That is why for f(x) that curved part is chosen.

Here on the graph of f(x) there is two sharp corner at x = 0 and x = 1. As we know no function is differentiable at the sharp corner. So f(x) is not differentiable at those two sharp corner.

f : R $ \to $ R, with f(0) = f(1) = 0
and f'(0) = 0
$ \because $ f(x) is differentiable and continuous
and f(0) = f(1) = 0

Applying Rolle’s theorem in [0, 1] for function f(x)
f'(c) = 0, c $ \in $ (0, 1)

Now again
$ \because $ f'(c) = 0, f'(0) = 0
again applying Rolles theorem in [0, c] for function f'(x)
f''(c₁) = 0 for some c₁ $ \in $ (0, c) $ \in $ (0, 1)

$\mathop {\lim }\limits_{x \to 0} {{x\left( {{e^{\left( {\sqrt {1 + {x^2} + {x^4}} - 1} \right)/x}} - 1} \right)} \over {\sqrt {1 + {x^2} + {x^4}} - 1}}$

= $\mathop {\lim }\limits_{x \to 0} {{x\left[ {{e^{{{\left( {\sqrt {1 + {x^2} + {x^4}} - 1} \right)\left( {\sqrt {1 + {x^2} + {x^4}} + 1} \right)} \over {x\left( {\sqrt {1 + {x^2} + {x^4}} + 1} \right)}}}} - 1} \right] \times \left( {\sqrt {1 + {x^2} + {x^4}} + 1} \right)} \over {\left( {\sqrt {1 + {x^2} + {x^4}} - 1} \right)\left( {\sqrt {1 + {x^2} + {x^4}} + 1} \right)}}$

= $\mathop {\lim }\limits_{x \to 0} {{x\left[ {{e^{{{\left( {1 + {x^2} + {x^4}} \right) - 1} \over {x\left( {\sqrt {1 + {x^2} + {x^4}} + 1} \right)}}}} - 1} \right] \times \left( {\sqrt {1 + {x^2} + {x^4}} + 1} \right)} \over {\left( {1 + {x^2} + {x^4} - 1} \right)}}$

= $\mathop {\lim }\limits_{x \to 0} {{\left[ {{e^{{{{x^2} + {x^4}} \over {x\left( {\sqrt {1 + 0 + 0} + 1} \right)}}}} - 1} \right] \times \left( {\sqrt {1 + 0 + 0} + 1} \right)} \over {x + {x^3}}}$

= $2\mathop {\lim }\limits_{x \to 0} {{\left[ {{e^{{{{x^2} + {x^4}} \over {2x}}}} - 1} \right]} \over {x + {x^3}}}$

= $2\mathop {\lim }\limits_{x \to 0} {{\left[ {{e^{{{x + {x^3}} \over 2}}} - 1} \right]} \over {{{x + {x^3}} \over 2} \times 2}}$

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

= 1

Note : As from formula, $\mathop {\lim }\limits_{x \to 0} {{\left[ {{e^{{{x + {x^3}} \over 2}}} - 1} \right]} \over {{{x + {x^3}} \over 2}}}$ = 1

Given, $f\left( x \right) = \left\{ {\matrix{ {{k_1}{{\left( {x - \pi } \right)}^2} - 1,} & {x \le \pi } \cr {{k_2}\cos x,} & {x > \pi } \cr } } \right.$

Differentiating one time,

$f'\left( x \right) = \left\{ {\matrix{ {2{k_1}\left( {x - \pi } \right),} & {x \le \pi } \cr { - {k_2}\sin x,} & {x > \pi } \cr } } \right.$

Differentiating one more time,

$f''\left( x \right) = \left\{ {\matrix{ {2{k_1},} & {x \le \pi } \cr { - {k_2}\cos x,} & {x > \pi } \cr } } \right.$

As f''(x) is differentiable so

f''($\pi $⁺) = f''($\pi $^-)

$ \Rightarrow $ -k₂(-1) = 2k₁

$ \Rightarrow $ 2k₁ = k₂

$ \therefore $ (k₁, k₂) = $\left( {{1 \over 2},1} \right)$

${x^2} - x - 2 = 0$

roots are 2 & $-$1 $ \Rightarrow $ $\alpha $ = 2 (given $\alpha$ is positive)

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

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

$ = \mathop {\lim }\limits_{x \to {2^ + }} {{\sqrt 2 \sin \left( {{{(x - 2)(x + 1)} \over 2}} \right)} \over {(x - 2)}}$

$ = {3 \over {\sqrt 2 }}$

$\mathop {\lim }\limits_{t \to x} {{{t^2}{f^2}(x) - {x^2}{f^2}(t)} \over {t - x}} = 0$

(Using L'Hospital's Rule)

$ \Rightarrow \mathop {\lim }\limits_{t \to x} {{2t{f^2}(x) - 2{x^2}f(t).f'(t)} \over 1} = 0$

$ \Rightarrow $ 2xf²(x) - 2x².f(x).f'(x) = 0

$ \Rightarrow $ 2xf(x){f(x) - xf'(x)} = 0

[x $ \ne $ 0, f(x) $ \ne $ 0 as given
function $f:\left( {0,\infty } \right) \to \left( {0,\infty } \right)$ only takes positive value as input and output]

$ \Rightarrow f(x) = xf'(x) $

$\Rightarrow {{f'(x)} \over {f(x)}} = {1 \over x}$

Integrating w.r.t x, we get

$ \Rightarrow ln\,f(x) = ln\,x + ln\,C$

$ \Rightarrow f(x) = Cx$

$ \because $ f(1) = e

$ \Rightarrow C = e;\,so\,f(x) = ex$

When f(x) = 1 = ex

$ \Rightarrow x = {1 \over e}$

$f\left( x \right) = \left\{ {\matrix{ {{\pi \over 4} + {{\tan }^{ - 1}}x,} & {x \in \left[ { - 1,1} \right]} \cr {{1 \over 2}\left( {x - 1} \right),} & {x > 1} \cr {{1 \over 2}\left( { - x - 1} \right),} & {x < - 1} \cr } } \right.$

At x = 1

L.H.L = $\mathop {\lim }\limits_{x \to {1^ - }} \left( {{\pi \over 4} + {{\tan }^{ - 1}}x} \right)$ = ${{\pi \over 4} + {\pi \over 4}}$ = ${{\pi \over 2}}$

f(1) = ${{\pi \over 4} + {{\tan }^{ - 1}}x}$ = ${{\pi \over 4} + {\pi \over 4}}$ = ${{\pi \over 2}}$

R.H.L = $\mathop {\lim }\limits_{x \to {1^ + }} \left( {{1 \over 2}\left( {x - 1} \right)} \right)$ = 0

As L.H.L $ \ne $ R.H.L so function is discontinuous $ \Rightarrow $ non differentiable.

At x = -1

L.H.L = $\mathop {\lim }\limits_{x \to - {1^ - }} \left( {{1 \over 2}\left( { - x - 1} \right)} \right)$ = ${{1 \over 2}\left( { - \left( { - 1} \right) - 1} \right)}$ = 0

f(-1) = ${\pi \over 4} + {\tan ^{ - 1}}\left( { - 1} \right)$ = ${\pi \over 4} - {\pi \over 4}$ = 0

R.H.L = $\mathop {\lim }\limits_{x \to - {1^ + }} \left( {{\pi \over 4} + {{\tan }^{ - 1}}x} \right)$

= ${\pi \over 4} + {\tan ^{ - 1}}\left( { - 1} \right)$ = ${\pi \over 4} - {\pi \over 4}$ = 0

As L.H.L = f(-1) = R.H.L so function is continuous.

$f'\left( x \right) = \left\{ {\matrix{ {{1 \over {1 + {x^2}}},} & {x \in \left[ { - 1,1} \right]} \cr {{1 \over 2},} & {x > 1} \cr { - {1 \over 2},} & {x < - 1} \cr } } \right.$

For differentiability at x = –1

L.H.D = ${ - {1 \over 2}}$

R.H.D. = ${{1 \over 2}}$

So, non differentiable at x = –1

L = $\mathop {\lim }\limits_{x \to a} {{{{\left( {a + 2x} \right)}^{{1 \over 3}}} - {{\left( {3x} \right)}^{{1 \over 3}}}} \over {{{\left( {3a + x} \right)}^{{1 \over 3}}} - {{\left( {4x} \right)}^{{1 \over 3}}}}}$

$= \mathop {\lim }\limits_{h \to 0} {{{{(a + 2(a + h))}^{1/3}} - {{(3(a + h))}^{1/3}}} \over {{{(3a + a + h)}^{1/3}} - {{(4(a + h))}^{1/3}}}}$

= $\mathop {\lim }\limits_{h \to 0} {{{{(3a)}^{1/3}}{{\left( {1 + {{2h} \over {3a}}} \right)}^{1/3}} - {{(3a)}^{1/3}}{{\left( {1 + {h \over a}} \right)}^{1/3}}} \over {{{(4a)}^{1/3}}{{\left( {1 + {h \over {4a}}} \right)}^{1/3}} - {{(4a)}^{1/3}}{{\left( {1 + {h \over a}} \right)}^{1/3}}}}$

= $\mathop {\lim }\limits_{h \to 0} \left( {{{{3^{1/3}}} \over {{4^{1/3}}}}} \right)\left[ {{{\left( {1 + {{2h} \over {9a}}} \right) - \left( {1 + {h \over {3a}}} \right)} \over {\left( {1 + {h \over {12a}}} \right) - \left( {1 + {h \over {3a}}} \right)}}} \right]$

$ = {\left( {{3 \over 4}} \right)^{1/3}}{{\left( {{2 \over 9} - {1 \over 3}} \right)} \over {\left( {{1 \over {12}} - {1 \over 3}} \right)}} = {\left( {{3 \over 4}} \right)^{1/3}}\left( {{{8 - 12} \over {3 - 12}}} \right)$

$ = {\left( {{3 \over 4}} \right)^{1/3}}\left( {{{ - 4} \over { - 9}}} \right) = {{{4^{1 - {1 \over 3}}}} \over {{3^{2 - {1 \over 3}}}}} = {{{4^{2/3}}} \over {{3^{5/3}}}}$

$ = {{{{(8 \times 2)}^{1/3}}} \over {{{(27 \times 9)}^{1/3}}}} = {2 \over 3}{\left( {{2 \over 9}} \right)^{1/3}}$

Here $\mathop {\lim }\limits_{x \to 0} \left| {{{1 - x + \left| x \right|} \over {\lambda - x + [x]}}} \right| = L$

Here L.H.L. $\mathop {\lim }\limits_{h \to 0^-} \left| {{{1 + h + h} \over {\lambda + h - 1}}} \right| = \left| {{1 \over {\lambda - 1}}} \right|$

R.H.L. = $\mathop {\lim }\limits_{h \to 0^+} \left| {{{1 - h + h} \over {\lambda + h + 0}}} \right| = \left| {{1 \over \lambda }} \right|$

$ \because $ Limit exists. Hence L.H.L. = R.H.L.

$ \Rightarrow $ $\left| {\lambda - 1} \right| = \left| \lambda \right|$

$ \Rightarrow $ $\lambda = {1 \over 2}$

$ \therefore $ L = ${1 \over {\left| \lambda \right|}}$ = 2

$\mathop {\lim }\limits_{x \to 0} {\left( {\tan \left( {{\pi \over 4} + x} \right)} \right)^{{1 \over x}}}$

This is 1^{$\infty $} form.

= ${e^{\mathop {\lim }\limits_{x \to 0} \left[ {\tan \left( {{\pi \over 4} + x} \right) - 1} \right] \times {1 \over x}}}$

= ${e^{\mathop {\lim }\limits_{x \to 0} \left[ {{{1 + \tan x} \over {1 - \tan x}} - 1} \right] \times {1 \over x}}}$

= ${e^{\mathop {\lim }\limits_{x \to 0} \left[ {{{2\tan x} \over {x\left( {1 - \tan x} \right)}}} \right]}}$

= ${e^{2\mathop {\lim }\limits_{x \to 0} \left[ {{{\tan x} \over x} \times {1 \over {\left( {1 - \tan x} \right)}}} \right]}}$

= ${e^{2\mathop {\lim }\limits_{x \to 0} \left[ {1 \times {1 \over {\left( {1 - 0} \right)}}} \right]}}$

= e²

Given function,
$f\left( x \right) = \left\{ {\matrix{ {a{e^x} + b{e^{ - x}},} & { - 1 \le x < 1} \cr {c{x^2},} & {1 \le x \le 3} \cr {a{x^2} + 2cx,} & {3 < x \le 4} \cr } } \right.$

For continuity at x = 1

$\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right)$

$ \Rightarrow $ $ae + b{e^{ - 1}} = c$

$ \Rightarrow $ b = ce - $a$e² .....(1)

For continuity at x = 3

$\mathop {\lim }\limits_{x \to {3^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {3^ + }} f\left( x \right)$

$ \Rightarrow $ 9c = 9a + 6c

$ \Rightarrow $ c = 3a .......(2)

Also given, f'(0) + f'(2) = e

$ \Rightarrow $ (ae^x – be^x )_x=0 + (2c^x)_x=2 = e

$ \Rightarrow $ a – b + 4c = e ........(3)

From (1), (2) & (3)

a – 3ae + ae² + 12a = e

$ \Rightarrow $ a(e² + 13 – 3e) = e

$ \Rightarrow $ a = ${e \over {{e^2} - 3e + 13}}$

A = $\mathop {\lim }\limits_{x \to 0} x\left[ {{4 \over x}} \right]$

= $\mathop {\lim }\limits_{x \to 0} x\left( {{4 \over x} - \left\{ {{4 \over x}} \right\}} \right)$

= $\mathop {\lim }\limits_{x \to 0} \left( {4 - \left\{ {{4 \over x}} \right\}} \right)$

= 4

Now, when x = $\sqrt {A + 1} $ = $\sqrt 5 $, f(x) = [x²]sin($\pi $x) is discontinuous at this non integer point.

But at x = 2, 3 and 5, f(x) is continuous.

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

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

= $\left( {a + 2} \right)$ + 1

= $\left( {a + 3} \right)$

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

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

= $\mathop {\lim }\limits_{x \to {0^ + }} {{1 + x - 1} \over x}$

= 1

And f(0) = b

As f(x) is continuous at x = 0, then

f(0^-) = f(0) = f(0⁺)

$ \Rightarrow $ $a + 3$ = b = 1

$ \therefore $ $a$ = -2 and b = 1

$ \therefore $ $a$ + 2b = -2 + 2 = 0

It is clear from graph that, slope of AC $>$ slope of CB

$ \Rightarrow $ ${{f\left( c \right) - f\left( a \right)} \over {c - a}}$ $>$ ${{f\left( b \right) - f\left( c \right)} \over {b - c}}$

$ \Rightarrow $ ${{f\left( c \right) - f\left( a \right)} \over {f\left( b \right) - f\left( c \right)}}$ $ > {{c - a} \over {b - c}}$

If we consider the case where f(x) is a constant function, then its derivative f'(x) is equal to 0 for all x in the interval (0,1).

Therefore, if we substitute this into the expressions provided in Options A, B and C, we would have :

Option A : |f(c) - f(1)| < |f'(c)| would become |constant - constant| < |0|, which is 0 < 0. This is not true.

Option B : |f(c) + f(1)| < (1 + c)|f'(c)| would become |constant + constant| < (1 + c)$ \times $0, which is a positive number < 0. This is not true.

Option C : |f(c) - f(1)| < (1 - c)|f'(c)| would become |constant - constant| < (1 - c)$ \times $0, which is 0 < 0. This is not true.

Hence, for the case where f(x) is a constant function, none of the options A, B and C are correct.

So, the correct answer would be Option D : None.

Given $\mathop {\lim }\limits_{x \to 0} {\left( {{{3{x^2} + 2} \over {7{x^2} + 2}}} \right)^{{1 \over {{x^2}}}}}$

Putting x = 0 we get 1^{$\infty $} form.

$ \therefore $ ${e^{\mathop {\lim }\limits_{x \to 0} {1 \over {{x^2}}}\left[ {{{3{x^2} + 2} \over {7{x^2} + 2}} - 1} \right]}}$

= ${e^{\mathop {\lim }\limits_{x \to 0} {1 \over {{x^2}}}\left[ {{{ - 4{x^2}} \over {7{x^2} + 2}}} \right]}}$

= e^-4/2

= e^-2 = ${1 \over {{e^2}}}$

The given function f : R $ \to $ R is satisfying as

$f(x + y) = f(x) + f(y) + f(x)f(y)$

So, $f'(x) = \mathop {\lim }\limits_{h \to 0} {{f(x + h) - f(x)} \over h}$

$ = \mathop {\lim }\limits_{h \to 0} {{f(x) + f(h) + f(x)f(h) - f(x)} \over h}$

$ = \mathop {\lim }\limits_{h \to 0} {{f(h)} \over h}(1 + f(x))$

$ \because $ $f(x) = xg(x) \Rightarrow g(x) = {{f(x)} \over x}$

$ \therefore $ $\mathop {\lim }\limits_{x \to 0} {{f(x)} \over x} = \mathop {\lim }\limits_{x \to 0} g(x) = 1$ (given)

$ \therefore $ $f'(x) = 1 + f(x)) \Rightarrow {{f'(x)} \over {1 + f(x)}} = 1$

$ \Rightarrow {\log _e}(1 + f(x)) = x + c$

$ \Rightarrow 1 + f(x) = {e^{x + c}}$

$ \Rightarrow f(x) = {e^{x + c}} - 1$

$ \because $ $f(0) = 0 \Rightarrow c = 0$

Therefore, $f(x) = {e^x} - 1$ is differentiable at every $x \in R$.

And $f'(x) = {e^x} \Rightarrow f'(0) = 1$

Now, $g(x) = f{{(x)} \over x} = {{{e^x} - 1} \over x}$ and if g(0) = 1

LHD (at x = 0) of

$g(x) = \mathop {\lim }\limits_{h \to 0} {{g(0 - h) - g(0)} \over { - h}}$

$ = \mathop {\lim }\limits_{h \to 0} {{{{{e^{ - h}} - 1} \over { - h}} - 1} \over { - h}}$

$ = \mathop {\lim }\limits_{h \to 0} {{{e^{ - h}} - 1 + h} \over {{h^2}}} = {1 \over 2}$

and, RHD (at x = 0) of

$g(x) = \mathop {\lim }\limits_{h \to 0} {{g(0 + h) - g(0)} \over h}$

$ = \mathop {\lim }\limits_{h \to 0} {{{e^h} - 1 - h} \over {{h^2}}} = {1 \over 2}$

So, if g(0) = 1, then g is differentiable at every x$ \in $R.

Given functions f : R $ \to $ R be defined

by f(x) = x³ $-$ x² + (x + 1) sin x and g : R $ \to $ R be an arbitrary function.

Now, let g is continuous at x = 1, then

$\mathop {\lim }\limits_{x \to {1^ - }} {{(fg)(x) - (fg)(1)} \over {x - 1}}$

$ = \mathop {\lim }\limits_{h \to 0} {{(fg)(1 - h) - (fg)(1)} \over {1 - h - 1}}$

$ \because $ $(fg)(x) = f(x)\,.\,g(x)$ (given)

$ = \mathop {\lim }\limits_{h \to 0} {{f(1 - h)\,.\,g(1 - h) - f(1)\,.\,g(1)} \over { - h}}$

$ = \mathop {\lim }\limits_{h \to 0} {{f(1 - h)\,.\,g(1)} \over { - h}}$

{$ \because $ f(1) = 0 and g is continuous at x = 1, so g(1 $-$ h) = g(1)}

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

$ = (1 + \sin 1)g(1)$

Similarly, $\mathop {\lim }\limits_{x \to {1^ + }} {{(fg)(x) - (fg)(1)} \over {x - 1}}$

$ = \mathop {\lim }\limits_{h \to 0} {{f(1 - h)\,.\,g(1)} \over h}$

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

$ = (1 + \sin 1)g(1)$

$ \because $ RHD and LHD of function fg at x = 1 is finitely exists and equal, so fg is differentiable at x = 1

Now, let (fg)(x) is differentiable at x = 1, so

$\mathop {\lim }\limits_{x \to {1^ - }} {{(fg)(x) - (fg)(1)} \over {x - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} {{(fg)(x) - (fg)(1)} \over {x - 1}}$

$ \Rightarrow \mathop {\lim }\limits_{x \to {1^ - }} {{f(x)g(x) - f(1)g(1)} \over {x - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} {{f(x)g(x) - f(1)g(1)} \over {x - 1}}$

$ \Rightarrow $ $\mathop {\lim }\limits_{x \to {1^ - }} {{f(x)g(x)} \over {x - 1}} = \mathop {\lim }\limits_{x \to {1^ + }} {{f(x)g(x)} \over {x - 1}}$ {$ \because $ f(1) = 0}

$ \Rightarrow $ $\mathop {\lim }\limits_{h \to 0} {{f(1 - h)\,g(1 - h)} \over { - h}} = \mathop {\lim }\limits_{h \to 0} {{f(1 + h)\,g(1 + h)} \over h}$

$ \Rightarrow $ $\mathop {\lim }\limits_{h \to 0} {{[{{(1 - h)}^2}( - h) + ( - h)\sin (1 + h)]g(1 + h)} \over { - h}}$

= $\mathop {\lim }\limits_{h \to 0} {{[{{(1 + h)}^2}(h) + (h)\sin (1 + h)]g(1 + h)} \over h}$

$\mathop {\lim }\limits_{h \to 0} \,[{(1 - h)^2} + \sin (1 - h)]g(1 - h) $

$= \mathop {\lim }\limits_{h \to 0} \,[{(1 + h)^2} + \sin (1 + h)]g(1 + h)$

It does not mean that g(x) is continuous or differentiable at x = 1.

But if g is differentiable at x = 1, then it must be continuous at x = 1 and so fg is differentiable at x = 1.

$ \begin{array}{r} \text {Given that, } \mathop {\lim }\limits_{x \to 0} \frac{1-\cos (1-\cos x)}{\sin ^4 x} \\ =\mathop {\lim }\limits_{x \to 0} \frac{1-\cos (1-\cos x)}{(1-\cos x)^2} \times \frac{(1-\cos x)^2}{x^4} \times \frac{x^4}{\sin ^4 x} \\ =\mathop {\lim }\limits_{x \to 0} \frac{1-\cos (1-\cos x)}{(1-\cos x)^2} \times \mathop {\lim }\limits_{x \to 0} \left(\frac{1-\cos x}{x^2}\right)^2 \\ \times \mathop {\lim }\limits_{x \to 0} \left(\frac{x}{\sin x}\right)^4 \end{array} $

$ \begin{aligned} & \text { As } x \rightarrow 0 \\ & (1-\cos x) \rightarrow 0 \\ & =\lim _{(1-\cos x) \rightarrow 0} \frac{1-\cos (1-\cos x)}{(1-\cos x)^2} \\ & \quad \times \lim _{x \rightarrow 0} \frac{(1-\cos x)^2}{x^2} \times \lim _{x \rightarrow 0}\left(\frac{x}{\sin x}\right)^4 \\ & =\frac{1}{2} \times\left(\frac{1}{2}\right)^2 \times(1)^2 \\ & \left\{\because \lim _{x \rightarrow 0} \frac{1-\cos x}{x^2}=\frac{1}{2} \text { and } \lim _{x \rightarrow 0} \frac{x}{\sin x}=1\right\} \\ & =\frac{1}{8} \end{aligned} $

$ \begin{aligned} &\text { Given that, }\\ &\begin{aligned} & f(x)=\left\{\begin{array}{cc} \frac{x}{|x|+2 x^2}, & x \neq 0 \\ k, & x=0 \end{array}\right. \\ & \mathrm{LHL}=\lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h) \end{aligned} \end{aligned} $

$ \begin{aligned} & =\lim _{h \rightarrow 0} \frac{-h}{|-h|+2(-h)^2}=\frac{-h}{h+2 h^2}=\frac{-h}{h(1+2 h)} \\ & =\lim _{h \rightarrow 0} \frac{-1}{1+2 h}=\frac{-1}{1+2 \times 0}=-1 \\ & \text { RHL }=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h) \\ & \quad=\lim _{h \rightarrow 0} \frac{h}{|h|+2 h^2}=\frac{h}{h(1+2 h)} \\ & \quad=\lim _{h \rightarrow 0} \frac{1}{1+2 h}=\frac{1}{1+2 \times 0}=1 \end{aligned} $

Also, $f(0)=k$

∵ LHL $\neq$ RHL, therefore given function is discontinuous for all values of $k$.

$ \begin{aligned} &\text { We have, }\\ &\begin{aligned} & \lim _{x \rightarrow k} \frac{\sin \left(2 \pi\left(([x])-\left[\frac{x}{k}\right]\right)-x\right)+\sin k}{x-k} \\ & =\lim _{x \rightarrow k} \frac{\sin (2 \pi m-x)+\sin k}{x-k} \\ & {\left[\because[x]-\left[\frac{x}{k}\right]=m \text { is an integer }\right]} \\ & =\lim _{x \rightarrow k} \frac{-\sin x+\sin k}{x-k} \\ & =\lim _{x \rightarrow k} \frac{-\cos x}{1}\left[\text { By } L^{\prime} \text { Hospital's rule }\right] \\ & =-\cos k \end{aligned} \end{aligned} $

We have,

$ \begin{aligned} f(x) & = \begin{cases}1+x, & 0 \leq x \leq 2 \\ 3-x, & 2< x \leq 3\end{cases} \\ f(f(x)) & = \begin{cases}1+f(x), & 0 \leq f(x) \leq 2 \\ 3-f(x), & 2< f(x) \leq 3\end{cases} \\ f f(x) & = \begin{cases}1+(1+x), & 0 \leq x \leq 1 \\ 3-(1+x), & 1< x \leq 2 \\ 1+(3-x), & 2< x \leq 3\end{cases} \\ f \circ f(x) & = \begin{cases}2+x, & 0 \leq x \leq 1 \\ 2-x, & 1< x \leq 2 \\ 4-x, & 2< x \leq 3\end{cases} \end{aligned} $

$\therefore f \circ f(x)$ is discontinuous at $x=1$, and

$ \begin{aligned} & x=2 \\ & \therefore \quad a=1, b=2 \\ & \quad 2 a+3 b=2(1)+3(2)=2+6=8 \end{aligned} $

We have,

$ \begin{aligned} & \lim _{x \rightarrow 0} \frac{1-\cos \left(x^2+\pi(x+2)\right.}{x^2} \\ & \lim _{x \rightarrow 0} \frac{1-\cos \left(2 \pi+\pi x+x^2\right)}{x^2} \\ & \lim _{x \rightarrow 0} \frac{1-\cos \left(\pi x+x^2\right)}{x^2} \\ & {[\because \cos (2 \pi+\theta)=\cos \theta]} \end{aligned} $

Applying 'L' Hospital rule

$ \begin{aligned} & \lim _{x \rightarrow 0} \frac{\sin \left(\pi x+x^2\right)(2 x+\pi)}{2 x} \\ = & \lim _{x \rightarrow 0} \sin \left(\pi x+x^2\right) \\ & \quad+\lim _{x \rightarrow 0} \frac{\pi \sin \left(\pi x+x^2\right)}{2 x} \\ = & 0+\frac{\pi}{2} \lim _{x \rightarrow 0} \frac{\sin \left(\pi x+x^2\right)}{x} \end{aligned} $

Again apply 'L' Hospital rule

$ \begin{aligned} & =\frac{\pi}{2} \lim _{x \rightarrow 0} \frac{\cos \left(\pi x+x^2\right)(2 x+\pi)}{1} \\ & =\frac{\pi}{2} \times \pi=\frac{\pi^2}{2} \end{aligned} $

Given,

$ f(x)=\left\{\begin{array}{ccc} \frac{1-\cos 4 x}{x^2} & , & \text { when } x<0 \\ a & , & x=0 \\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}-4}} & , & x>0 \end{array}\right. $

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

$ \therefore \quad(\mathrm{LHL})_{X=0}=(\mathrm{RHL})_{X=0}=f(0) $

Now, (LHL)

$ \begin{aligned} x=0 & =\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} \frac{1-\cos 4 x}{x^2} \\ & =\lim _{h \rightarrow 0} \frac{1-\cos 4(0-h)}{(0-h)^2} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & =\lim _{h \rightarrow 0} \frac{1-\cos 4 h}{h^2}=\lim _{h \rightarrow 0} \frac{2 \sin ^2 2 h}{h^2} \\ & =2 \lim _{h \rightarrow 0}\left(\frac{\sin 2 h}{2 h}\right)^2 \times 4 \\ & =2 \times(1)^2 \times 4=8 \end{aligned}\\ &\text { From Eq. (i), we have }\\ &\begin{aligned} & (\mathrm{LHL})_x=0=f(0) \Rightarrow 8=a \\ \Rightarrow \quad & a=8 \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { We have, }\\ &\lim _{x \rightarrow 0} \frac{\log (1+x)^{1+x}}{x^2}-\frac{1}{x}=k \end{aligned} $

$ \begin{array}{lc} \Rightarrow & \mathop {\lim }\limits_{x \to 0} \frac{(1+x) \log (1+x)-x}{x^2}=k \\ \Rightarrow & \mathop {\lim }\limits_{x \to 0} \frac{\log (1+x)+1-1}{2 x}=k \\ \Rightarrow & \quad \text { [using L' Hospital Rule] } \\ \Rightarrow & \frac{1}{2} \mathop {\lim }\limits_{x \to 0} \frac{\log (1+x)}{x}=k \\ \Rightarrow & \frac{1}{2} \times 1=k \quad\left[\because \mathop {\lim }\limits_{x \to 0} \frac{\log (1+x)}{x}=1\right] \\ \Rightarrow & k=\frac{1}{2} \quad \therefore \quad 12 k=12 \times \frac{1}{2}=6 \end{array} $

$ \begin{aligned} & \text { We have, } \\ & f(x) \text { is continuous at } x=1 \\ & \therefore \quad k=\lim _{x \rightarrow 1} \frac{(9 x-1)(\sqrt{x}-1)}{3 x^2+2 x-5} \\ & \Rightarrow \quad k=\lim _{x \rightarrow 1} \frac{9 x^{3 / 2}-9 x-\sqrt{x}+1}{3 x^2+2 x-5} \\ & \Rightarrow \quad k=\lim _{x \rightarrow 1} \frac{9\left(\frac{3}{2} x^{1 / 2}\right)-9-\frac{1}{2 \sqrt{x}}}{6 x+2} \\ & \Rightarrow \quad k=\frac{\frac{27}{2}-9-\frac{1}{2}}{6+2}=\frac{13-9}{8}=\frac{4}{8}=\frac{1}{2} \end{aligned} $

Let $f(x)=\frac{2 x-1}{3 x-4},[1,2]$

Since, $f(x)$ is not defined at

$ x=\frac{4}{3} \in[1,2] $

So, Lagrange's theorem is not applicable on $f(x)$ on $[1,2]$

$\mathop {\lim }\limits_{n \to \infty } {1 \over n}\left\{ {{1 \over {n + 1}} + {2 \over {n + 2}} + .... + {{3n} \over {n + 3n}}} \right\}$

$ = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^{3n} {{1 \over n}\left( {{r \over {n + r}}} \right) = \int_0^3 {{x \over {1 + x}}dx} } $

$ = \int_0^3 {\left( {1 - {1 \over {1 + x}}} \right)dx} $

$ = [x - \log (1 + x)]_0^3$

$ = 3 - \log 4 = 3 - 2\log 2$

From f(x) = 5 - | x - 2 |
maximum value of f(x) is at x = 2

From g(x) = | x + 1 |
minimum value of g(x) is at x = -1

$ \therefore $ $\alpha \beta $ = - 2

$ \Rightarrow $ $\mathop {\lim }\limits_{x \to 2} {{(x - 1)(x - 2)(x - 3)} \over {(x - 2)(x - 4)}}$

$ \Rightarrow $ ${{(2 - 1)(2 - 3)} \over {(2 - 4)}} = {1 \over 2}$