Limits, Continuity and Differentiability

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

$ \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)

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

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

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

Let $x = {\pi \over 2} + x;\,\,y \to 0$

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

$=$ $\mathop {\lim }\limits_{y \to 0} {{ - \tan {y \over 2}2{{\sin }^2}{y \over 2}} \over {\left( { - 8} \right).{{{y^3}} \over 8}.8}}$

$ = \mathop {\lim }\limits_{y \to 0} {1 \over {32}}{{\tan {y \over 2}} \over {\left( {{y \over 2}} \right)}}.{\left[ {{{\sin \,y/2} \over {y/2}}} \right]^2} = {1 \over {32}}$

$\mathop {\lim }\limits_{x \to a} {{f\left( a \right)g'\left( x \right) - g\left( a \right)f'\left( x \right)} \over {g'\left( x \right) - f'\left( x \right)}}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ (By $L'$ Hospital rule)

$\mathop {\lim }\limits_{x \to a} {{k\,\,g'\left( x \right) - k\,\,f'\left( x \right)} \over {g'\left( x \right) - f'\left( x \right)}} = 4$

$\therefore$ $k=4.$

$\lim {{\sqrt {1 - \cos \,2x} } \over {\sqrt 2 x}} \Rightarrow \lim {{\sqrt {1 - \left( {1 - 2\,{{\sin }^2}\,x} \right)} } \over {\sqrt 2 x}}$

$\mathop {\lim }\limits_{x \to 0} {{\sqrt {2\,{{\sin }^2}\,x} } \over {\sqrt {2x} }} \Rightarrow \mathop {\lim }\limits_{x \to 0} {{\left| {\sin x} \right|} \over x}$

The limit of above does not exist as

$LHS = - 1 \ne RHL = 1$

Let a is a rational number other than $0,$ in

$\left[ { - 5,5} \right],$ then

$f\left( a \right) = a$ and $\mathop {\lim }\limits_{x \to a} \,\,f\left( x \right) = - a$

[ As in the immediate neighbourhood of a rational -

number, we find irrational numbers ]

$\therefore$ $f(x)$ is not continuous at any rational number

If a is irrational number, then

$\,f\left( a \right) = - a$ and $\mathop {\lim }\limits_{x \to a} \,f\left( x \right) = a$

$\therefore$ $f(x)$ is not continuous at any irrational number clearly

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

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

$f\left( {x + y} \right) = f\left( x \right) \times f\left( y \right)$

Differeniate with respect to $x,$ treating $y$ as constant

$f'\left( {x + y} \right) = f'\left( x \right)f\left( y \right)$

Putting $x=0$ and $y=x$, we get

$f'\left( x \right) = f'\left( 0 \right)f\left( x \right);$

$ \Rightarrow f'\left( 5 \right) = 3f\left( 5 \right) = 3 \times 2 = 6.$

$\mathop {\lim }\limits_{x \to \infty } {\left( {{{{x^2} + 5x + 3} \over {{x^2} + x + 2}}} \right)^x}$

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

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

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

$ = {e^{\mathop {\lim }\limits_{x \to \infty } {{4{x^2} + x} \over {{x^2} + x + 2}}}}$

[ As $\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \lambda x} \right)^{{1 \over x}}} = {e^\lambda }$ ]

$ = {e^{\mathop {\lim }\limits_{x \to \infty } {{4 + {1 \over x}} \over {1 + {1 \over x} + {2 \over {{x^2}}}}}}} = {e^4}$

Since $\mathop {\lim }\limits_{x \to 0} \left[ x \right]$ does not exist, hence the required limit does not exist.

$\mathop {\lim }\limits_{x \to 1} {{\sqrt {f\left( x \right)} - 1 } \over {\sqrt x - 1}}\,\,\left( {{0 \over 0}} \right)$ form using $L'$ Hospital's rule

$ = \mathop {\lim }\limits_{x \to 1} {{{1 \over {2\sqrt {f\left( x \right)} }}f'\left( x \right)} \over {1/2\sqrt x }}$

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

To find the value of $f(x) - g(x)$ at $x = \frac{3}{2}$, we need to use the given conditions and properties of differentiable functions.

First, we are told that:

$f''(x) - g''(x) = 0$

This implies that:

$f''(x) = g''(x)$

Since the second derivatives of both functions are equal, their difference, $f'(x) - g'(x)$, must be a linear function. Let’s denote it as:

$f'(x) - g'(x) = k$

We'll find the constant $k$ using the initial conditions of the derivatives:

$f'(1) = 2$

$g'(1) = 4$

Thus,

$f'(1) - g'(1) = 2 - 4 = -2$

Therefore,

$f'(x) - g'(x) = -2$

Integrating the above result, we get:

$f(x) - g(x) = -2x + C$

To determine the constant $C$, we use the values of the functions at $x = 2$:

$f(2) = 3$

$g(2) = 9$

Thus,

$f(2) - g(2) = 3 - 9 = -6$

Therefore,

$-2 \cdot 2 + C = -6$

$-4 + C = -6$

$C = -2$

So the expression for $f(x) - g(x)$ is:

$f(x) - g(x) = -2x - 2$

We need to find $f\left( \frac{3}{2} \right) - g\left( \frac{3}{2} \right)$:

$f\left( \frac{3}{2} \right) - g\left( \frac{3}{2} \right) = -2 \cdot \frac{3}{2} - 2$

$= -3 - 2$

$= -5$

Thus, the value of $f(x) - g(x)$ at $x = \frac{3}{2}$ is -5.

Given,

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

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

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

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

           $\left[ {As\,f'\left( x \right) = \mathop {\lim }\limits_{x \to 2} {{f\left( x \right) - f\left( 2 \right)} \over {x - 2}}} \right]$

= $f\left( 2 \right) - 2f'\left( x \right)$

= 4 - 2 $ \times $ 4

= - 4