Limits, Continuity and Differentiability

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)

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

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

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

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