Limits, Continuity and Differentiability

$\mathop {\lim }\limits_{x \to \pi /4} {{{{\cot }^3}x - \tan x} \over {\cos \left( {x + {\pi \over 4}} \right)}}$

= $\mathop {\lim }\limits_{x \to {\pi \over 4}} {{{1 \over {{{\tan }^3}x}} - \tan x} \over {\cos \left( {x + {\pi \over 4}} \right)}}$

= $\mathop {\lim }\limits_{x \to {\pi \over 4}} {{1 - {{\tan }^4}x} \over {\left( {{{\tan }^3}x} \right)\cos \left( {x + {\pi \over 4}} \right)}}$

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

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

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

= $\mathop {\lim }\limits_{x \to {\pi \over 4}} {{2\left( {1 - {{\tan }^2}x} \right)} \over {\left( {{{\cos x - \sin x} \over {\sqrt 2 }}} \right)}}$

= $\mathop {\lim }\limits_{x \to {\pi \over 4}} {{2\sqrt 2 \left( {1 - {{{{\sin }^2}x} \over {{{\cos }^2}x}}} \right)} \over {\left( {\cos x - \sin x} \right)}}$

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

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

= $\mathop {\lim }\limits_{x \to {\pi \over 4}} {{2\sqrt 2 \left( {\cos x + \sin x} \right)} \over {{{\cos }^2}x}}$

= $\mathop {\lim }\limits_{x \to {\pi \over 4}} {{2\sqrt 2 \left( {\cos {\pi \over 4} + \sin {\pi \over 4}} \right)} \over {{{\cos }^2}{\pi \over 4}}}$

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

= ${{2\sqrt 2 \left( {{2 \over {\sqrt 2 }}} \right)} \over {{1 \over 2}}}$

= 8

Other Method :

$\mathop {\lim }\limits_{x \to \pi /4} {{{{\cot }^3}x - \tan x} \over {\cos \left( {x + {\pi \over 4}} \right)}}$

This is in ${0 \over 0}$ form so L' Hospital rule is applicable.

= $\mathop {\lim }\limits_{x \to {\pi \over 4}} {{3{{\cot }^3}x\left( { - \cos e{c^2}x} \right) - {{\sec }^2}x} \over { - \sin \left( {x + {\pi \over 4}} \right)}}$

= ${{3 \times 1\left( { - {{\left( {\sqrt 2 } \right)}^2}} \right) - {{\left( {\sqrt 2 } \right)}^2}} \over { - 1}}$

= ${{ - 6 - 2} \over { - 1}}$ = 8

$\mathop {\lim }\limits_{x \to 0} {{x{{\tan }^2}2x} \over {\tan 4x{{\sin }^2}x}} = \mathop {\lim }\limits_{x \to 0} {{x\left( {{{{{\tan }^2}2x} \over {4{x^2}}}} \right)4{x^2}} \over {\left( {{{\tan 4x} \over {4x}}} \right)4x\left( {{{{{\sin }^2}x} \over {{x^2}}}} \right){x^2}}} = 1$

f(x) = sin$\left| x \right| - \left| x \right|$ + 2(x $-$ $\pi $) cosx

$ \because $  sin$\left| x \right|$ $-$ $\left| x \right|$ is differentiable function at c = 0

$ \therefore $  k = $\phi $

R.H.L. $=$ $\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right) + {{\left( {\left| x \right| - \sin \left( {x\left[ x \right]} \right)} \right)}^2}} \over {{x^2}}}$

(as x $ \to $ 0⁺ $ \Rightarrow $  [x] $=$ 0)

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

$=$ $\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right)} \over {\left( {\pi {{\sin }^2}x} \right)}} + 1 = \pi + 1$

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

(as x $ \to $ 0^$-$ $ \Rightarrow $  [x] $=$ $-$1)

$\mathop {\lim }\limits_{x \to {0^ + }} {{\tan \left( {\pi {{\sin }^2}x} \right)} \over {\pi {{\sin }^2}x}}\,.\,{{\pi {{\sin }^2}x} \over {{x^2}}} + {\left( { - 1 + {{\sin x} \over x}} \right)^2} \Rightarrow \pi $

R.H.L.  $ \ne $  L.H.L.

$\left| {f\left( x \right)} \right| = \left\{ {\matrix{ 1 & , & { - 2 \le x < 0} \cr {1 - {x^2}} & , & {0 \le x < 1} \cr {{x^2} - 1} & , & {1 \le x \le 2} \cr } } \right.$

and  $f\left( {\left| x \right|} \right) = {x^2} - 1,x \in \left[ { - 2,2} \right]$

Hence  $g(x) = \left\{ {\matrix{ {{x^2}} & , & {x \in \left[ { - 2,0} \right]} \cr 0 & , & {x \in \left[ {0,1} \right)} \cr {2\left( {{x^2} - 1} \right)} & , & {x \in \left[ {1,2} \right]} \cr } } \right.$

It is not differentiable at x = 1

f : ($-$ 1, 1) $ \to $ R

f(x) = max {$-$ $\left| x \right|, - \sqrt {1 - {x^2}} $}


Non-derivable at 3 points in ($-$1, 1)

$\mathop {\lim }\limits_{x \to {1^ + }} {{\left( {1 - \left| x \right| + \sin \left| {1 - x} \right|} \right)\sin \left( {{\pi \over 2}\left[ {1 - x} \right]} \right)} \over {\left| {1 - x} \right|\left[ {1 - x} \right]}}$

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

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

$f\left( x \right) = \left\{ {\matrix{ {8 + 2x,} & { - 4 \le x \le - 2} \cr {{x^2},} & { - 2 \le x \le - 1} \cr {\left| x \right|,} & { - 1 < x < 1} \cr {{x^2},} & {1 \le x \le 2} \cr {8 - 2x,} & {2 < x \le 4} \cr } } \right.$


f(x) is not differentiable at

x = $\left\{ { - 2, - 1,0,1,2} \right\}$

$ \Rightarrow $  S = {$-$2, $-$ 1, 0, 1, 2}

$\mathop {\lim }\limits_{x \to {0^ - }} {{x\left( {\left[ x \right] + \left| x \right|} \right)\sin \left[ x \right]} \over {\left| x \right|}}$

$ = \mathop {\lim }\limits_{h \to 0} {{\left( {0 - h} \right)\left( {\left[ {0 - h} \right] + \left| {0 - h} \right|} \right)\sin \left[ {0 - h} \right]} \over {\left| {0 - h} \right|}}$

$ = \mathop {\lim }\limits_{h \to 0} {{ - h\left( { - 1 + h} \right)\sin \left( { - 1} \right)} \over h}$

$ = \mathop {\lim }\limits_{h \to 0} \left( { - 1 + h} \right)\sin \left( 1 \right) = - \sin 1$

$\mathop {\lim }\limits_{y \to 0} {{\sqrt {1 + \sqrt {1 + {y^4}} } - \sqrt 2 } \over {{y^4}}}$

If you put y = 0 at ${{\sqrt {1 + \sqrt {1 + {y^4}} } - \sqrt 2 } \over {{y^4}}}$ it is in ${0 \over 0}$ form. So we can use L' Hospital's Rule.

= $\mathop {\lim }\limits_{y \to 0} {{{1 \over {2\sqrt {1 + \sqrt {1 + {y^4}} } }} \times \left( {{1 \over {2\sqrt {1 + {y^4}} }}} \right) \times 4{y^3}} \over {4{y^3}}}$

= $\mathop {\lim }\limits_{y \to 0} {1 \over {2\sqrt {1 + \sqrt {1 + {y^4}} } }} \times {1 \over {2\sqrt {1 + {y^4}} }}$

= ${1 \over {4\sqrt 2 }}$

Checking

if f(x) is continuous at x = 1 :

f(1^$-$) = 5

f(1) = 5

f(1⁺) = a + b

if f(x) is continuous at x = 1,

then

f(1^$-$) = f(1) = f(1⁺)

$ \Rightarrow $  5 = 5 = a + b

$ \therefore $  a + b = 5 . . . . . . . . (1)

checking if f(x) is continuous at x = 3 :

f(3^$-$) = a + 3b

f(3) = b + 15

f(3⁺) = b + 15

if   f(x) = is continuous at x = 3

then,

f(3^$-$) = f(3) = f(3⁺)

$ \Rightarrow $  a + 3b = b + 15 = b + 15

$ \Rightarrow $  a + 2b = 15 . . . . . (2)

checking if f(x) is continuous at x = 5 :

f(5^$-$) = b + 25

f(5) = 30

f(5⁺) = 30

if f(x) is continuous at x = 5 then,

f(5^$-$) = f(5) = f(5⁺)

$ \Rightarrow $   b + 25 = 30 = 30

$ \Rightarrow $   b = 5

By putting this value in equation (2), we get,

a + 2(5) = 15

$ \Rightarrow $  a = 5

when a = 5 and b = 5 then equation (1)

a + b = 5 does not satisfy.

$ \therefore $  f is not continuous for any value of a and b.

Given,

$\mathop {lim}\limits_{x \to 0} \,{{{{\left( {27 + x} \right)}^{{1 \over 3}}} - 3} \over {9 - {{\left( {27 + x} \right)}^{{2 \over 3}}}}}$

=   $\mathop {\lim }\limits_{x \to 0} \,{{3\left[ {{{\left( {1 + {x \over {27}}} \right)}^{{1 \over 3}}} - 1} \right]} \over {9\left[ {1 - {{\left( {1 + {x \over {27}}} \right)}^{{2 \over 3}}}} \right]}}$

=   $\mathop {\lim }\limits_{x \to 0} \,\,{{\left( {1 + {x \over {3 \times 27}}} \right) - 1} \over {3\left[ {1 - \left( {1 + {{2x} \over {3 \times 27}}} \right)} \right]}}$

=   $\mathop {\lim }\limits_{x \to 0} \,\,{{{x \over {81}}} \over {3\left( { - {{2x} \over {81}}} \right)}} = - {1 \over 6}$

If the function is continuous at x = 0, then
$\mathop {\lim }\limits_{x \to 0} $ f(x) will exist and f(0) = $\mathop {\lim }\limits_{x \to 0} $ f(x)

Now, $\mathop {\lim }\limits_{x \to 0} $ f(x) = $\mathop {\lim }\limits_{x \to 0} \left( {{1 \over x} - {{k - 1} \over {{e^{2x}} - 1}}} \right)$

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

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

= $\mathop {\lim }\limits_{x \to 0} \left[ {{{\left( {3 - k} \right)x + {{4{x^2}} \over {2!}} + {{8{x^3}} \over {3!}} + ...} \over {\left( {2{x^2} + {{4{x^3}} \over {2!}} + {{8{x^3}} \over {3!}} + ....} \right)}}} \right]$

For the limit to exist, power of x in the numerator should be greater than or equal to the power of x in the denominator. Therefore, coefficient of x in numerator is equal to zero

$ \Rightarrow $  3 $-$ k = 0

$ \Rightarrow $  k = 3

So the limit reduces to

$\mathop {\lim }\limits_{x \to 0} {{\left( {{x^2}} \right)\left( {{4 \over {2!}} + {{8x} \over {3!}} + ...} \right)} \over {\left( {{x^2}} \right)\left( {2 + {{4x} \over {2!}} + {{8{x^2}} \over {3!}} + ...} \right)}}$

= $\mathop {\lim }\limits_{x \to 0} {{{4 \over {2!}} + {{8x} \over {3!}} + ...} \over {2 + {{4x} \over {2!}} + {{8{x^2}} \over {3!}} + ...}}$ = 1

Hence, f(0) = 1

Given,

$\mathop {lim}\limits_{x \to {0^ + }} \,\,x\,\left( {\left[ {{1 \over x}} \right] + \left[ {{2 \over x}} \right]} \right. + $ $\left. {\,.\,.\,.\,.\,.\, + \left[ {{{15} \over x}} \right]} \right)$

as we know that

${1 \over x} = \left[ {{1 \over x}} \right] + \left\{ {{1 \over x}} \right\}$

$ \Rightarrow \,\,\,\,\left[ {{1 \over x}} \right] = {1 \over x} - \left\{ {{1 \over x}} \right\}$

$ = \,\,\,\,\mathop {\lim }\limits_{x \to {0^ + }} \,\,x\left[ {{1 \over x} - \left\{ {{1 \over x}} \right\} + {2 \over 2} - \left\{ {{2 \over x}} \right\} + ........{{15} \over x} - \left\{ {{{15} \over x}} \right\}} \right]$

$ = \,\,\,\,\mathop {\lim }\limits_{x \to {0^ + }} \,\left[ {x.{1 \over x} + x.{2 \over x} + .....x.{{15} \over x}} \right]$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \,\left[ {x.\left\{ {{1 \over 2}} \right\} + .. + x.\left\{ {{{15} \over x}} \right\}} \right]$

We know $\left\{ {{1 \over x}} \right\}$ is fractional part of ${1 \over x}.$

So, the range of $\,\left\{ {{1 \over x}} \right\}$ is $0 \le \left\{ {{1 \over x}} \right\} < 1$

So, $\mathop {lim}\limits_{x \to {0^ + }} \,\,x\,\left\{ {{1 \over x}} \right\} = 0.$ (finite no) $=0$

Similarly $\mathop {\lim }\limits_{x \to {0^ + }} x.\left\{ {{2 \over x}} \right\} = 0$

$ = \,\,\,\,\left( {1 + 2 + ... + 15} \right) - \left( {0 + 0...} \right)$

$ = \,\,\,\,{{15 \times 16} \over 2}$

$ = \,\,\,\,120$

Check differtiability at x = $\pi $ and x = 0

at x = 0 :

We have L. H. D = $\mathop {\lim }\limits_{h \to 0} {{f\left( {0 - h} \right) - f\left( 0 \right)} \over { - h}}$

= $\mathop {\lim }\limits_{h \to 0} {{\left| { - h - \pi } \right|\left( {{e^{\left| { - h} \right|}} - 1} \right)\sin \left| h \right| - 0} \over { - h}} = 0$

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

$ = \mathop {\lim }\limits_{h \to 0} {{\left| {h - \pi } \right|\left( {{e^{\left| h \right|}} - 1} \right)\sin \left| h \right| - o} \over h}$

= 0

$\therefore,\,\,$ LHD = RHD

Therefore, function is differentiable at x = $\pi $.

at x = $\pi $ :

L. H. D = $\mathop {\lim }\limits_{h \to 0} {{f\left( {\pi - h} \right) - f\left( \pi \right)} \over { - h}}$

= $\mathop {\lim }\limits_{h \to 0} {{\left| {\pi - h - \pi } \right|\left( {{e^{\left| {\pi - h} \right|}} - 1} \right)\sin \left| {\pi - h} \right| - 0.} \over { - h}}$

= 0

RH. D = $\mathop {\lim }\limits_{h \to 0} {{f\left( {\pi + h} \right) - f\left( \pi \right)} \over h}$

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

= 0

$\therefore\,\,\,$ L. H. D = R H D

Therefore, function is differentiable at x = $\pi $,

Since, the function f(x) is differentiable at all the points including 0 and $\pi $.

So, f(x) is differentiable everywhere.

Therefore, set S is empty set.

S = $\phi $

$ \because $    f(x) has extremum values at x = 1, and x = 2

$ \because $  f'(1) = 0 and f'(2) = 0

As, f(x) is a polynomial of degree 4.

Suppose f(x) = Ax⁴ + Bx³ + cx² + Dx + E

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

$ \Rightarrow $$\,\,\,$$\mathop {\lim }\limits_{x \to 0} \left( {{{A{x^4} + B{x^3} + C{x^2} + Dx + E} \over {{x^2}}} + 1} \right)$ = 3

$ \Rightarrow $   $\mathop {\lim }\limits_{x \to 0} $ $\left( {A{x^2} + Bx + C + {D \over x} + {E \over {{x^2}}} + 1} \right)$ = 3

As limit has finite value, so D = 0 and E = 0

Now A(0)² + B(0) + C + 0 + 0 + 1 = 3

$ \Rightarrow $   c + 1 = 3 $ \Rightarrow $ c = 2

f'(x) = 4Ax³ + 3Bx² + 2Cx + D

f'(1) = 0 $ \Rightarrow $ 4A(1) + 3B(1) + 2C(1) + D = 0

$ \Rightarrow $$\,\,\,$ 4A + 3B = $-$ 4     . . . .(1)

f'(2) = 0 $ \Rightarrow $ 4A(8) + 3B(4) + 2C(2) + D = 0

$ \Rightarrow $ 8A + 3B = $-$ 2      . . . . .(2)

From equations (1) and (2), we get

A = ${1 \over 2}$ and B = $-$ 2

So, f(x) = ${{{x_4}} \over 2} - 2{x^3} + 2x{}^2$

Therefore, f($-$ 1) = ${{{{( - 1)}^4}} \over 2} - 2{( - 1)^3} + 2{( - 1)^2}$

= ${1 \over 2} + 2 + 2 = {9 \over 2}$

Hence f($-$1) = ${9 \over 2}$

Since f(x) is continuous at x = 2.

$\therefore\,\,\,$$\mathop {\lim }\limits_{x \to 2} $ f(x) = f(2)

$ \Rightarrow $  $\mathop {\lim }\limits_{x \to 2} $ ${\left( {x - 1} \right)^{{1 \over {2 - x}}}}$ = k    (${{1^\infty }}$  form)

$ \therefore $  ${{e^l}}$ = k

Where $l$ = $\mathop {\lim }\limits_{x \to 2} $(x $-$ $l$ $-$ 1) $ \times $ ${1 \over {2 - x}}$ = $\mathop {\lim }\limits_{x \to 2} {{x - 2} \over {2 - x}}$

= $\mathop {\lim }\limits_{x \to 2} \left( {{{x - 2} \over {x - 2}}} \right)$

$ \Rightarrow $    k = e^$-$1

Let, L = $\mathop {\lim }\limits_{x \to 0} {{\left( {x\tan 2x - 2x\tan x} \right)} \over {{{\left( {1 - \cos 2x} \right)}^2}}}$ = $\mathop {\lim }\limits_{x \to 0} K$ (say)

$ \Rightarrow $ K = ${{x\left[ {{{2\tan x} \over {1 - {{\left( {\tan x} \right)}^2}}}} \right] - 2x\tan x} \over {{{\left( {1 - \left( {1 - 2{{\sin }^2}x} \right)} \right)}^2}}}$

= ${{2x\tan x - \left[ {2x\tan x - 2x{{\tan }^3}x} \right]} \over {4{{\sin }^4}x \times (1 - {{\tan }^2}x)}}$

= ${{2x{{\tan }^3}x} \over {4{{\sin }^4}x \times \left( {1 - {{\tan }^2}x} \right)}}$

= ${{2x{{\tan }^3}x} \over {4{{\sin }^4}x \times \left( {{{{{\cos }^2}x - {{\sin }^2}x} \over {{{\cos }^2}x}}} \right)}}$

= ${{2x{{{{\sin }^3}x} \over {{{\cos }^3}x}}} \over {4{{\sin }^4}x \times \left( {{{{{\cos }^2}x - {{\sin }^2}x} \over {{{\cos }^2}x}}} \right)}}$

$ \Rightarrow $K = ${x \over {2\sin x \times ({{\cos }^2}x - {{\sin }^2}x)\cos x}}$

$\therefore\,\,\,$ L = $\mathop {\lim }\limits_{x \to 0} $ ${x \over {2\sin x}} \times \mathop {\lim }\limits_{x \to 0} {1 \over {\cos x({{\cos }^2}x - {{\sin }^2}x)}}$

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

= ${1 \over 2}$

S = {($\lambda $, $\mu $) $ \in $ R $ \times $ R : f(t) = $\left( {\left| \lambda \right|{e^{\left| t \right|}} - \mu } \right)$ sin $\left( {2\left| t \right|} \right),$ t $ \in $ R

f(t) = $\left( {\left| \lambda \right|{e^{\left| t \right|}} - \mu } \right)\sin \left( {2\left| t \right|} \right)$

= $\left\{ {\matrix{ {\left( {\left| \lambda \right|{e^t} - \mu } \right)\sin 2t,} & {t > 0} \cr {\left( {\left| \lambda \right|{e^{ - t}} - \mu } \right)\left( { - \sin 2t} \right),} & {t < 0} \cr } } \right.$

f'(t) = $\left\{ {\matrix{ {\left( {\left| \lambda \right|{e^t}} \right)\sin 2t + \left( {\left| \lambda \right|{e^t} - \mu } \right)\left( {2\cos 2t} \right),\,\,t > 0} \cr {\left| \lambda \right|{e^{ - t}}\sin 2t + \left( {\left| \lambda \right|{e^{ - t}} - \mu } \right)\left( {-2\cos 2t} \right),t < 0} \cr } } \right.$

As, f(t) is differentiable

$ \therefore $  LHD = RHD at t = 0

$ \Rightarrow $   $\left| \lambda \right|$ . sin2(0) + $\left( {\left| \lambda \right|{e^0} - \mu } \right)$2cos(0)

      = $\left| \lambda \right|{e^{ - 0}}\,$ . sin 2(0) $-$ 2 cos (0) $\left( {\left| \lambda \right|{e^{ - 0}} - \mu } \right)$

$ \Rightarrow $$\,\,\,$ 0 + $\left( {\left| \lambda \right| - \mu } \right)$2 = 0 $-$ 2$\left( {\left| \lambda \right| - \mu } \right)$

$ \Rightarrow $  4$\left( {\left| \lambda \right| - \mu } \right)$ = 0

$ \Rightarrow $  $\left| \mu \right|$ = $\mu $

So, S $ \equiv $ ($\lambda $, $\mu $) = {$\lambda $ $ \in $ R & $\mu $ $ \in $ [0, $\infty $)}

Therefore set S is subset of R $ \times $ [0, $\infty $)

$f(x)=\left(\frac{4}{5}\right)^{\frac{\tan 4 x}{\tan 5 x}}$

$\lim\limits_{x \rightarrow \frac{x}{2}}\left(\frac{4}{5}\right)^{\frac{\tan 4 x}{\tan 5 x}}=k+\frac{2}{5}$

$\Rightarrow \lim\limits_{x \rightarrow \frac{x}{2}}\left(\frac{4}{5}\right)^{\tan 4 x \cot 5 x}=k+\frac{2}{5}$

$\Rightarrow\left(\frac{4}{5}\right)^{\lim\limits_{x \rightarrow \frac{x}{2}}(\tan 4 x \cdot \cot (5 x))}=k+\frac{2}{5}$

$\Rightarrow\left(\frac{4}{5}\right)^{0 \times \cos \left(2 x+\frac{x}{2}\right)}=k+\frac{2}{5}$

$\Rightarrow\left(\frac{4}{5}\right)^0=k+\frac{2}{5}$

$\Rightarrow k+\frac{2}{5}=1 \Rightarrow k=1-\frac{2}{5} \Rightarrow k=\frac{3}{5}$

Given,

      $\mathop {\lim }\limits_{x \to 3} $ ${{\sqrt {3x} - 3} \over {\sqrt {2x - 4} - \sqrt 2 }}$

Here if you put x = 3 in ${{\sqrt {3x} - 3} \over {\sqrt {2x - 4 - \sqrt 2 } }}$

you will get ${0 \over 0}$ form.

So, we can apply L' Hospital rule

$\therefore\,\,\,$ $\mathop {\lim }\limits_{x \to 3} {{\sqrt {3x} - 3} \over {\sqrt {2x - 4} - \sqrt 2 }}$

= $\mathop {\lim }\limits_{x \to 3} $ ${{\sqrt 3 .{1 \over {2\sqrt x }}} \over {{2 \over {2\sqrt {2x - 4} }}}}$ (applying L' Hospital rule)

= ${{\sqrt 3 .{1 \over {2\sqrt 3 }}} \over {{1 \over {\sqrt 6 - 4}}}}$

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

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

$\mathop {\lim }\limits_{x \to {\pi \over 2}} {{\cot x - \cos x} \over {{{\left( {\pi - 2x} \right)}^3}}}$

= $\mathop {\lim }\limits_{x \to {\pi \over 2}} {1 \over 8}.{{\cos x\left( {1 - \sin x} \right)} \over {\sin x{{\left( {{\pi \over 2} - x} \right)}^3}}}$

Put ${{\pi \over 2} - x}$ = t

$ \Rightarrow $ x $ \to $ ${{\pi \over 2}}$

t $ \to $ 0

= $\mathop {\lim }\limits_{t \to 0} {1 \over 8}.{{\cos \left( {{\pi \over 2} - t} \right)\left( {1 - \sin \left( {{\pi \over 2} - t} \right)} \right)} \over {\sin \left( {{\pi \over 2} - t} \right){{\left( {{\pi \over 2} - {\pi \over 2} + t} \right)}^3}}}$

= $\mathop {{1 \over 8}\lim }\limits_{t \to 0} {{\sin t\left( {1 - \cos t} \right)} \over {\cos t{{\left( t \right)}^3}}}$

= $\mathop {{1 \over 8}\lim }\limits_{t \to 0} {{\sin t\left( {2{{\sin }^2}{t \over 2}} \right)} \over {\cos t{{\left( t \right)}^3}}}$

= $\mathop {{1 \over 4}\lim }\limits_{t \to 0} {{\sin t\left( {{{\sin }^2}{t \over 2}} \right)} \over {\cos t{{\left( t \right)}^3}}}$

= $\mathop {\lim }\limits_{t \to 0} \left( {{{\sin t} \over t}} \right){\left( {{{\sin {t \over 2}} \over {{t \over 2}}}} \right)^2}{1 \over {\cos t}}{1 \over 4}$

= ${1 \over 4} \times {1 \over 4}$

= ${1 \over {16}}$

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

$ = \mathop {\lim }\limits_{x \to 0} {{{{\left( {2{{\sin }^2}x} \right)}^2}} \over {2x\left( {x + {{{x^3}} \over 3} + {{2{x^5}} \over {15}} + ....} \right) - x\left( {2x + {{{2^3}{x^3}} \over 3} + 2.{{{2^5}{x^5}} \over {15}}+...} \right)}}$

$ = \mathop {\lim }\limits_{x \to 0} {{4{{\left( {x - {{{x^3}} \over {3!}} + {{{x^5}} \over {5!}} - ....} \right)}^4}} \over {{x^4}\left( {{2 \over 3} - {8 \over 3}} \right) + {x^6}\left( {{4 \over {15}} - {{64} \over {15}}} \right) + ....}}$

$ = \mathop {\lim }\limits_{x \to 0} {{4{{\left( {1 - {{{x^2}} \over {3!}} + {{{3^4}} \over {5!}} - ....} \right)}^4}} \over { - 2 + {x^2}\left( { - {{60} \over {15}}} \right) + ....}}$

(By dividing numerator and denominator by x⁴)

$ = {{4{{\left( {1 - 0} \right)}^4}} \over { - 2 + 0}}$

$ = {4 \over { - 2}}$

$=$ $-$ 2

f(x) is continuous at x = 1

$ \therefore $    ${{2{{\left( 1 \right)}^2}} \over a} = a$

$ \Rightarrow $   a² = 2

$ \Rightarrow $   a = $ \pm $ $\sqrt 2 $

Also f(x) is continuous at x = $\sqrt 2 $

$ \therefore $   a = ${{2{b^2} - 4b} \over {2\sqrt 2 }}$

Now, when a = $\sqrt 2 ,$ then

4 = 2b² $-$ 4b

$ \Rightarrow $   b² $-$ 2b = 2

$ \Rightarrow $   b² $-$ 2b $-$ 2 = 0

$ \Rightarrow $   b = ${{2 \mp \sqrt {4 + 4.2} } \over 2}$

= 1 $ \pm $ $\sqrt 3 $

$ \therefore $   (a, b) = $\left( {\sqrt 2 ,1 \pm \sqrt 3 } \right)$

When a = $-$ $\sqrt 2 $, then

$-$ $\sqrt 2 $ = ${{2{b^2} - 4b} \over {2\sqrt 2 }}$

$ \Rightarrow $   $-$ 4 = 2b² $-$ 4b

$ \Rightarrow $   b² $-$ 2b + 2 = 0

$ \therefore $    b = ${{2 \pm \sqrt {4 - 8} } \over 2}$ = 1 $ \pm $ i

As  b $ \in $ Real number so,

b = 1 $ \pm $ i  is not accepted.

$ \therefore $   (a, b) = $\left( {\sqrt 2 ,1 + \sqrt 3 } \right)$   or  $\left( {\sqrt 2 ,1 - \sqrt 3 } \right)$

As   f(x) is differentiable at x = 1

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

$ \Rightarrow $   $-$1 $=$ a + cos^$-$1(1 + b)

$ \Rightarrow $   cos^$-$1 (1 + b) $=$ $-$1 $-$ a . . . . . .(1)

As   f(x) is differentiable, so,

2 . H . D $=$ R . H . D

Here, L . H . D $=$ $\mathop {\lim }\limits_{h \to 0} $ ${{f\left( {1 - h} \right) - f\left( 1 \right)} \over { - h}}$

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

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

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

$=$ $-$ 1

R. H. D $ = \mathop {\lim }\limits_{x \to 0} {{f\left( {1 + h} \right) - f\left( 1 \right)} \over h}$

$ = \mathop {\lim }\limits_{h \to 0} {{a + {{\cos }^{ - 1}}\left( {1 + h + b} \right) - \left[ {a + {{\cos }^{ - 1}}\left( {1 + b} \right)} \right]} \over h}$

$ = \mathop {\lim }\limits_{h \to 0} {{{{\cos }^{ - 1}}\left( {1 + h + b} \right) - {{\cos }^{ - 1}}\left( {1 + b} \right)} \over h}\left[ {{0 \over 0}\,\,form\left. \, \right]} \right.$

$ = \mathop {\lim }\limits_{h \to 0} {{ - 1} \over {\sqrt {1 - {{\left( {1 + h + b} \right)}^2}} }}$ [ Using L' Hospital Rule]

$ = {{ - 1} \over {\sqrt {1 - {{\left( {1 + b} \right)}^2}} }}$

$ \therefore $   $ - 1 = {{ - 1} \over {\sqrt {1 - {{\left( {1 + b} \right)}^2}} }}$

$ \Rightarrow 1 - {\left( {1 + b} \right)^2} = 1$

$ \Rightarrow $  ${\left( {1 + b} \right)^2} = 0$

$ \Rightarrow $   $b = - 1$

putting value of b in equation (1), we get,

${\cos ^{ - 1}}\left( {1 - 1} \right) = - 1 - a$

$ \Rightarrow $   ${\pi \over 2} = - 1 - a$

$ \Rightarrow $   $a = - 1 - {\pi \over 2}$

$ \therefore $   ${a \over b} = {{ - 1 - {\pi \over 2}} \over { - 1}} = 1 + {\pi \over 2} = {{\pi + 2} \over 2}$

Given,

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

So,   $\mathop {\lim }\limits_{x \to \propto } {\left( {1 + {a \over x} - {4 \over {{x^2}}}} \right)^{2x}}\,\left[ {{1^ \propto }\,\,\,form} \right]$

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

$ = {e^{\mathop {\lim }\limits_{x \to \propto } \left( {2a - {8 \over x}} \right)}}$

$ = {e^{2a}}$

$ \therefore $   e^2a = e³

$ \therefore $    2a = 3

$ \Rightarrow $   a $=$ ${3 \over 2}$

$\ln \,P = \mathop {\lim }\limits_{x \to {0^ + }} {1 \over {2x}}\ln \left( {1 + {{\tan }^2}\sqrt x } \right)$

$\mathop {\lim }\limits_{x \to {0^ + }} {1 \over x}\ln \left( {\sec \sqrt x \,\,\,\,\,\,} \right)$

Applying $L$ Hospital's rule :

$ = \mathop {\lim }\limits_{x \to {0^ + }} {{\sec \sqrt x \tan \sqrt x } \over {\sec \sqrt x .2\sqrt x }}$

$ = \mathop {\lim }\limits_{x \to {0^ + }} {{\tan \sqrt x } \over {2\sqrt x }}$

$ = {1 \over 2}$

$g\left( x \right) = f\left( {f\left( x \right)} \right)$

In the neighbourhood of $x=0,$

$f\left( x \right) = \left| {\log 2 - \sin \,x} \right| = \left( {\log 2 - \sin x} \right)$

$\therefore$ $g\left( x \right) = \left| {\log 2 - \sin \left. {\left| {\log 2 - \sin x} \right|} \right|} \right.$

$ = \left( {\log 2 - \sin \left( {\log 2 - \sin x} \right)} \right)$

$\therefore$ $g(x)$ is differentiable

and $g'\left( x \right) = - \cos \left( {\log 2 - \sin x} \right)\left( { - \cos x} \right)$

$ \Rightarrow g'\left( 0 \right) = \cos \left( {\log 2} \right)$

Multiply and divide by $x$ in the given expression, we get

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

$ = \mathop {\lim }\limits_{x \to 0} {{2{{\sin }^2}x} \over {{x^2}}}.{{3 + \cos x} \over 1}.{x \over {\tan \,4x}}$

$ = 2\mathop {\lim }\limits_{x \to 0} {{{{\sin }^2}x} \over {{x^2}}}.\mathop {\lim }\limits_{x \to 0} 3 + \cos x.\mathop {\lim }\limits_{x \to 0} {x \over {\tan 4x}}$

$ = 2.4{1 \over 4}\mathop {\lim }\limits_{x \to 0} {{4x} \over {\tan 4x}} = 2.4.{1 \over 4} = 2$

Since $g(x)$ is differentiable, -

it will be continuous at $x=3$

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

$2k = 3m + 2\,\,\,\,\,...\left( 1 \right)$

Also $g(x)$ is differentiable at $x=0$

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

${K \over {2\sqrt {3 + 1} }} = m$

$k=4m$ $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$

Solving $(1)$ and $(2)$, we get

$m = {2 \over 5},\,\,k = {8 \over 5}$

$\therefore$ $k+m=2$

Consider $\mathop {\lim }\limits_{x \to 0} {{\sin \left( {\pi {{\cos }^2}x} \right)} \over {{x^2}}}$

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

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

$\left[ \, \right.$ As $\sin \left( {\pi - \theta } \right) = \sin \theta $ $\left. \, \right]$

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

$ = \mathop {\lim }\limits_{x \to 0} 1 \times \pi {\left( {{{\sin x} \over x}} \right)^2} = \pi $

Multiply and divide by $x$ in the given expression, we get

$\mathop {\lim }\limits_{x \to 0} {{\left( {1 - \cos 2x} \right)} \over {{x_2}}}{{\left( {3 + \cos x} \right)} \over 1}.{x \over {\tan \,4x}}$

$ = \mathop {\lim }\limits_{x \to 0} {{2{{\sin }^2}x} \over {{x^2}}}.{{3 + \cos x} \over 1}.{x \over {\tan 4x}}$

$ = 2\mathop {\lim }\limits_{x \to 0} {{{{\sin }^2}x} \over {{x^2}}}.\mathop {\lim }\limits_{x \to 0} 3 + \cos x.\mathop {\lim }\limits_{x \to 0} {x \over {\tan 4x}}$

$ = 2.4{1 \over 4}\mathop {\lim }\limits_{x \to 0} {{4x} \over {\tan 4x}} = 2.4.{1 \over 4} = 2$

$f\left( x \right) = \left| {x - 2} \right| = \left\{ {\matrix{ {x - 2\,\,\,,} & {x - 2 \ge 0} \cr {2 - x\,\,\,,} & {x - 2 \le 0} \cr } } \right.$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left\{ {\matrix{ {x - 2\,\,\,,} & {x \ge 2} \cr {2 - x\,\,\,,} & {x \le 2} \cr } } \right.$

Similarly, $f\left( x \right) = \left| {x - 5} \right| = \left\{ {\matrix{ {x - 5\,\,\,,} & {x \ge 5} \cr {5 - x\,\,\,,} & {x \le 5} \cr } } \right.$

$\therefore$ $f\left( x \right) = \left| {x - 2} \right| + \left| {x - 5} \right|$

$ = \left\{ {x - 2 + 5 - x = 3,2 \le x \le 5} \right\}$

Thus $\,\,\,\,\,\,\,\,\,f\left( x \right) = 3,2 \le x \le 5$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,f'\left( x \right) = 0,2 < x < 5$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,f'\left( 4 \right) = 0$

$\therefore$ Statement $1$ is true.



As $f\left( 2 \right) = 0 + \left| {2 - 5} \right| = 3$

and $f\left( 5 \right) = \left| {5 - 2} \right| + 0 = 3$

$\therefore$ Statement - $2$ is also true but not a correct explanation for statement $1.$

Let $f\left( x \right) = \left[ x \right]\cos \left( {{{2x - 1} \over 2}} \right)$

Doubtful points are $x = n,n \in I$

$L.H.L$ $ = \mathop {\lim }\limits_{x \to {n^ - }} \left[ x \right]\cos \left( {{{2x - 1} \over 2}} \right)\pi $

$ = \left( {n - 1} \right)\cos \left( {{{2n - 1} \over 2}} \right)\pi = 0$

( As $\left[ x \right]$ is the greatest integer function )

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

Now, value of the function at $x = n$ is $f(n)=0$

Since, $L.H.L.$ $ = R.H.L. = f\left( n \right)$

$\therefore$ $f\left( x \right) = \left[ x \right]\cos \left( {{{2x - 1} \over 2}} \right)$ is continuous for every real $x.$

$L.H.L. = \mathop {\lim }\limits_{x \to 0} f\left( x \right) = \mathop {\lim }\limits_{h \to 0} {{\sin \left\{ {\left( {p + 1} \right)\left( { - h} \right)} \right\} - \sin \left( { - h} \right)} \over { - h}}$

$ = \mathop {\lim }\limits_{h \to 0} {{ - \sin \left( {p + 1} \right)h} \over { - h}} + {{\sin \left( { - h} \right)} \over { - h}} = p + 1 + 1 = p + 2$

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

$ = \mathop {\lim }\limits_{h \to 0} {1 \over {\left( {\sqrt {1 + h} + 1} \right)}} = {1 \over 2}$

and $f\left( 0 \right) = q \Rightarrow p = - {3 \over 2},q = {1 \over 2}$

$\mathop {\lim }\limits_{x \to 2} {{\sqrt {1 - \cos \left\{ {2\left( {x - 2} \right)} \right\}} } \over {x - 2}}$

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

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

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

Thus $L.H.L. \ne R.H.L.$

Hence, $\mathop {\lim }\limits_{x \to 2} {{\sqrt {1 - \cos \left\{ {2\left( {x - 2} \right)} \right\}} } \over {x - 2}}\,\,$ does not exist.

$f(x)$ is a positive increasing function.

$\therefore$ $0 < f\left( x \right) < f\left( {2x} \right) < f\left( {3x} \right)$

$ \Rightarrow 0 < 1 < {{f\left( {2x} \right)} \over {f\left( x \right)}} < {{f\left( {3x} \right)} \over {f\left( x \right)}}$

$ \Rightarrow \mathop {\lim }\limits_{x \to \infty } 1 \le \mathop {\lim }\limits_{x \to \infty } {{f\left( {2x} \right)} \over {f\left( x \right)}} \le \mathop {\lim }\limits_{x \to \infty } {{f\left( {3x} \right)} \over {f\left( x \right)}}$

By Sandwich Theorem.

$ \Rightarrow \mathop {\lim }\limits_{x \to \infty } {{f\left( {2x} \right)} \over {f\left( x \right)}} = 1$

Given that $f\left( x \right) = x\left| x \right|\,\,$ and $\,\,g\left( x \right) = \sin x$

So that go

$f\left( x \right) = g\left( {f\left( x \right)} \right)$

$ = g\left( {x\left| x \right|} \right) = \sin x\left| x \right|$

$ = \left\{ {\matrix{ {\sin \left( { - {x^2}} \right),} & {if\,\,\,x < 0} \cr {\sin \left( {{x^2}} \right),} & {if\,\,\,x \ge 0} \cr } } \right.$

$ = \left\{ {\matrix{ { - \sin \,{x^2},} & {if\,\,\,x < 0} \cr {\sin \,\,{x^2},} & {if\,\,\,x \ge 0} \cr } } \right.$

$\therefore$ $\left( {go\,f} \right)'\,\,\left( x \right) = \left\{ {\matrix{ { - 2x\,\,\cos \,{x^2},\,\,\,\,if\,\,\,\,x < 0} \cr {2x\,\cos \,{x^2},\,\,\,if\,\,\,\,x \ge 0} \cr } } \right.$

Here we observe

$L\left( {gof} \right)'\left( 0 \right) = 0 = R\left( {gof} \right)'\left( 0 \right)$

$ \Rightarrow $ go $f$ is differentiable at $x=0$

and $\left( {go\,f} \right)'$ is continuous at $x=0$

Now $\left( {go\,f} \right)''\left( x \right) = \left\{ {\matrix{ { - 2\cos {x^2} + 4{x^2}\sin {x^2},x < 0} \cr {2\cos {x^2} - 4{x^2}\sin {x^2},x \ge 0} \cr } } \right.$

Here $L\left( {gof} \right)''\left( 0 \right) = - 2$ and $R\left( {go\,f} \right)''\left( 0 \right) = 2$

As $L{\left( {go\,f} \right)^{''}}\left( 0 \right) \ne R\left( {go\,f} \right)''\,\,\left( 0 \right)$

$ \Rightarrow go\,f\left( x \right)$ is not twice differentiable at $x=0.$

$\therefore$ Statement - $1$ is true but statement $-2$ is false.

We have $f\left( x \right) = \left\{ {\matrix{ {\left( {x - 1} \right)\sin \left( {{1 \over {x - 1}}} \right),} & {if\,\,x \ne 1} \cr {0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} & {if\,\,x = 1} \cr } } \right.$

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

$ = \mathop {\lim }\limits_{h \to 0} {{h\,\sin {1 \over h} - 0} \over h} = \mathop {\lim }\limits_{h \to 0} \,\,\sin {1 \over h} = a$ finite number

Let this finite number be $l$

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

$ = \mathop {\lim }\limits_{h \to 0} {{ - h\,\sin \left( {{1 \over { - h}}} \right)} \over { - h}} = \mathop {\lim }\limits_{h \to 0} \,\,\sin \left( {{1 \over { - h}}} \right)$

$ = - \mathop {\lim }\limits_{h \to 0} \sin \left( {{1 \over h}} \right) = - (\,$ a finite number $\,)$ $=-l$

Thus $Rf'\left( 1 \right) \ne Lf'\left( 1 \right)$

$\therefore$ $f$ is not differentiable at $x=1$

Also, $f'\left( 0 \right) = \sin {1 \over {\left( {x - 1} \right)}} - {{x - 1} \over {{{\left( {x - 1} \right)}^2}}}\cos {\left. {\left( {{1 \over {x - 1}}} \right)} \right]_{x = 0}}$

$ = - \sin 1 + \cos \,1$

$\therefore$ $f$ is differentiable at $x=0$

$f\left( x \right) = \min \left\{ {x + 1,\left| x \right| + 1} \right\}$

$ \Rightarrow f\left( x \right) = x + 1\,\forall \,x \in R$



Hence, $f(x)$ is differentiable everywhere for all $x \in R.$

Given, $f\left( x \right) = {1 \over x} - {2 \over {{e^{2x}} - 1}}$

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

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

$\therefore$ using, $L$'Hospital rule

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

$ = \mathop {\lim }\limits_{x \to 0} {{4{e^{2x}}} \over {4x{e^{2x}} + 2{e^{2x}} + 2{e^{2x}}}}\,\,$ $\left[ {\,\,} \right.$ ${0 \over 0}$ form $\left. {\,\,} \right]$

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

$f\left( x \right) = \left\{ {\matrix{ {{x \over {1 - x}},} & {x < 0} \cr {{x \over {1 + x}},} & {x \ge 0} \cr } } \right.$

$ \Rightarrow f'\left( x \right) = \left\{ {\matrix{ {{x \over {{{\left( {1 - x} \right)}^2}}},} & {x < 0} \cr {{x \over {{{\left( {1 + x} \right)}^2}}}} & {x \ge 0} \cr } } \right.$

$\therefore$ $f'\left( x \right)$ exist at everywhere.

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

$\,\,\,\,\,\,\,\,\,\left| {f'\left( x \right)} \right| = \mathop {\lim }\limits_{h \to 0} \left| {{{f\left( {x + h} \right) - f\left( x \right)} \over h}} \right| \le \mathop {\lim }\limits_{h \to 0} \left| {{{{{\left( h \right)}^2}} \over h}} \right|$

$ \Rightarrow \left| {f'\left( x \right)} \right| \le 0 \Rightarrow f'\left( x \right) = 0$

$ \Rightarrow f\left( x \right) = $ constant

As $f\left( 0 \right) = 0 \Rightarrow f\left( 1 \right) = 0$

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

As function is differentiable so it is continuous as it

is given that $\mathop {\lim }\limits_{h \to 0} {{f\left( {1 + h} \right)} \over h} = 5$ and hence $f(1)=0$

Hence $f'(1)$ $ = \mathop {\lim }\limits_{h \to 0} {{f\left( {1 + h} \right)} \over h} = 5$

Given limit $ = \mathop {\lim }\limits_{x \to \alpha } {{1 - \cos \,a\left( {x - \alpha } \right)\left( {x - \beta } \right)} \over {{{\left( {x - \alpha } \right)}^2}}}$

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

$ = \mathop {\lim }\limits_{x \to \alpha } {2 \over {{{\left( {x - \alpha } \right)}^2}}} \times {{{{\sin }^2}\left( {a{{\left( {x - \alpha } \right)\left( {x - \beta } \right)} \over 2}} \right)} \over {{{{a^2}{{\left( {x - \alpha } \right)}^2}{{\left( {x - \beta } \right)}^2}} \over 4}}} \times {{{a^2}{{\left( {x - \alpha } \right)}^2}{{\left( {x - \beta } \right)}^2}} \over 4}$

$ = {{{a^2}{{\left( {\alpha - \beta } \right)}^2}} \over 2}.$

We know that $\mathop {\lim }\limits_{x \to \infty } \left( {1 + x{1 \over x}} \right) = e$

$\therefore$ $\mathop {\lim }\limits_{x \to \infty } {\left( {1 + {a \over x} + {b \over {{x^2}}}} \right)^{2x}} = {e^2}$

$ \Rightarrow \mathop {\lim }\limits_{x \to \infty } {\left[ {\left( {1 + {a \over x} + {b \over {{x^2}}}} \right)\left( {{1 \over {{a \over x} + {b \over {{x^2}}}}}} \right)} \right]^{2x\left( {{a \over x} + {b \over {{x^2}}}} \right)}} = {e^2}$

$ \Rightarrow {e^{\mathop {\lim }\limits_{x \to \infty } 2\left[ {a + {b \over x}} \right]}} = {e^2} \Rightarrow {e^{2a}} = e{}^2 \Rightarrow a = 1$

and $b \in R$

$f\left( x \right) = {{1 - \tan x} \over {4x - \pi }}$ is continuous in $\left[ {0,{\pi \over 2}} \right]$

$\therefore$ $f\left( {{\pi \over 4}} \right) = \mathop {\lim }\limits_{x \to {\pi \over 4}} f\left( x \right) = \mathop {\lim }\limits_{x \to {\pi \over 4}} \, + f\left( x \right) = \mathop {\lim }\limits_{h \to 0} f\left( {{\pi \over 4} + h} \right)$

$ = \mathop {\lim }\limits_{h \to 0} {{1 - \tan \left( {{\pi \over 4} + h} \right)} \over {4\left( {{\pi \over 4} + h} \right) - \pi }},h > 0 = \mathop {\lim }\limits_{h \to 0} {{1 - {{1 + \tan \,h} \over {1 - \tan \,h}}} \over {4h}}$

$ = \mathop {\lim }\limits_{h \to 0} \,{{ - 2} \over {1 - \tan \,h}}.{{\tan \,h} \over {4h}} = {{ - 2} \over 4} = - {1 \over 2}$

$f\left( 0 \right) = 0;\,\,f\left( x \right) = x{e^{ - \left( {{1 \over {\left| x \right|}} + {1 \over x}} \right)}}$

$R.H.L.\,\,$ $\,\,\,\mathop {\lim }\limits_{h \to 0} \left( {0 + h} \right){e^{ - 2/h}}$

$ = \,\mathop {\lim }\limits_{h \to 0} {h \over {{e^{2/h}}}} = 0$

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

therefore, $f(x)$ is continuous,

$R.H.D=$ $\,\,\,\mathop {\lim }\limits_{h \to 0} {{\left( {0 + h} \right){e^{ - \left( {{1 \over h} + {1 \over h}} \right)}} - 0} \over h} = 0$

$L.H.D.$ $\,\,\, = \mathop {\lim }\limits_{h \to 0} {{\left( {0 - h} \right){e^{ - \left( {{1 \over h} - {1 \over h}} \right)}} - 0} \over { - h}} = 1$

therefore, $L.H.D. \ne R.H.D.$

$f(x)$ is not differentiable at $x=0.$

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

(by $L'$ Hospital rule)

$\mathop {\lim }\limits_{x \to 0} {{{1 \over {3 + x}} - {{ - 1} \over {3 - x}}} \over 1} = K$

$\therefore$ ${2 \over 3} = K$