Maths M.C.Q (1 Marks)

and $\mathop {\lim }\limits_{x \to 1 + } \,\,\frac{1}{{|\,\,1 - x\,\,|}} = \mathop {\lim }\limits_{h \to 0} \frac{1}{{1 + h - 1}} = \infty $

Hence $\mathop {\lim }\limits_{x \to 1} \,\frac{1}{{|\,\,1 - x\,\,|}} = \infty .$

$ = \mathop {\lim }\limits_{n \to \infty } \frac{{{n^3}\,\left( {4 + \frac{4}{n} + \frac{1}{{{n^2}}}} \right)}}{{{n^3}\left( {1 + \frac{5}{n} + \frac{5}{{{n^2}}} - \frac{2}{{{n^3}}}} \right)}} = 4$

$\mathop {\lim }\limits_{x \to \,0 + } \,f(x) = \mathop {\lim }\limits_{h \to \,0} \left( {\frac{{{e^{1/h}} - 1}}{{{e^{1/h}} + 1}}} \right) $

$= \mathop {\lim }\limits_{h \to \,0} \frac{{{e^{1/h}}\left( {1 - \frac{1}{{{e^{1/h}}}}} \right)}}{{{e^{1/h}}\left( {1 + \frac{1}{{{e^{1/h}}}}} \right)}} = 1$

Similarly $\mathop {\lim }\limits_{x \to \,0 - } f(x) = - 1$. 

Hence limit does not exist.

Aliter : Apply $L-$ Hospital’s rule

$\mathop {\lim }\limits_{x \to 0} \frac{{\sin x - x + \frac{{{x^3}}}{6}}}{{{x^5}}} = \mathop {\lim }\limits_{x \to 0} \frac{{\cos x - 1 + \frac{{3{x^2}}}{6}}}{{5{x^4}}}$

$ = \mathop {\lim }\limits_{x \to 0} \frac{{ - \sin x + \frac{{6x}}{6}}}{{20{x^3}}} = \mathop {\lim }\limits_{x \to 0} \,\frac{{ - \cos x + 1}}{{60{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{{120\,x}}$

$ = \mathop {\lim }\limits_{x \to 0} \frac{{\cos x}}{{120}} = \frac{1}{{120}}.$

Aliter : Apply $L-$ Hospital’s rule, 

$\mathop {\lim }\limits_{x \to 0} \,\left[ {\frac{{x - \log \,(1 + x)}}{{{x^2}}}} \right]$

$ = \mathop {\lim }\limits_{x \to 0} \frac{{1 - \frac{1}{{1 + x}}}}{{2x}}$

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

Note : Students should remember that

$\mathop {\lim }\limits_{n \to \infty } \,\frac{{\sum n}}{{{n^2}}} = \frac{1}{2},\,\,\,\mathop {\lim }\limits_{n \to \infty } \,\frac{{\sum {n^2}}}{{{n^3}}} = \frac{1}{3}$ 

and $\mathop {\lim }\limits_{n \to \infty } \,\frac{{\sum {n^3}}}{{{n^4}}} = \frac{1}{4}.$

$ = f(a)g'(a) - g(a)f'(a) = 2 \times 2 - ( - 1)\,(1) = 5.$

$ = \mathop {\lim }\limits_{y \to 0} \,\left[ {\frac{x}{y}\,\left\{ {\frac{{\cos x - \cos \,(x + y)}}{{\cos \,(x + y)\,\cos x}}} \right\}} \right] + \mathop {\lim }\limits_{y \to 0} \sec \,(x + y)$

$ = \mathop {\lim }\limits_{y \to 0} \,\left[ {\frac{{x\sin \,\left( {x + \frac{y}{2}} \right)}}{{\cos \,(x + y)\,.\,\,\cos x}}\,.\,\frac{{\sin \,\left( {\frac{y}{2}} \right)}}{{\,\,\,\frac{y}{2}}}} \right] + \sec x$

$= x\ tan\ x\ sec\ x + sec\ x = sec\ x\ (x\ tan\ x+1).$

Aliter : Apply $L-$ Hospital’s rule,

$\mathop {\lim }\limits_{y \to 0} \,\frac{{(x + y)\,\sec \,(x + y) - x\,\sec x}}{y}$

$ = \mathop {\lim }\limits_{y \to 0} \,\frac{{(x + y)\,\sec \,(x + y)\tan \,(x + y) + \sec \,(x + y) - 0}}{1}$

{Differentiating w.r.t.$y$ assuming $x$ as constant}

$ = x\sec x\tan x + \sec x. = \sec x(x\tan x + 1)$

$\mathop {\lim }\limits_{x \to 1} \,\frac{{1 - \sqrt x }}{{{{({{\cos }^{ - 1}}x)}^2}}} = \mathop {\lim }\limits_{y \to 0} \,\frac{{1 - \sqrt {\cos y} }}{{{y^2}}}$

Now rationalizing it, we get

$\mathop {\lim }\limits_{y \to 0} \,\frac{{(1 - \cos y)}}{{{y^2}(1 + \sqrt {\cos y} )}}$

$ = \mathop {\lim }\limits_{y \to 0} \,\frac{{1 - \cos y}}{{{y^2}}}\,.\,\mathop {\lim }\limits_{y \to 0} \,\frac{1}{{1 + \sqrt {\cos y} }} $

$= \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}.$

Therefore, given function $ = f'(a) + kf'(e) = 1$

$ \Rightarrow \,\,\frac{1}{a} + \frac{k}{e} = 1\,\, $

$\Rightarrow \,\,k = e\,\left( {\frac{{a - 1}}{a}} \right)$

Aliter : Apply $ L-$ Hospital’s rule to find both the limits.

$ = \mathop {\lim }\limits_{x \to \infty } \,{\left( {1 + \frac{1}{{x + 1}}} \right)^{x + 3}}$

$ = \mathop {\lim }\limits_{x \to \infty } \,{\left[ {{{\left( {1 + \frac{1}{{x + 1}}} \right)}^{x + 1}}} \right]^{\frac{{\,(x + 3)}}{{(x + 1)}}}} = e$

$\left\{ \because \,\,\underset{x\to \infty }{\mathop{\lim }}\,\,{{\left( 1+\frac{1}{x+1} \right)}^{x+1}}=e \right.$

and $\mathop {\lim }\limits_{x \to \infty } \frac{{\,(x + 3)}}{{(x + 1)}} = \left. {\mathop {\lim }\limits_{x \to \infty } \frac{{\,\left\{ {1 + (3/x)} \right\}}}{{\left\{ {1 + (1/x)} \right\}}} = 1} \right\}$.

Apply $L-$ Hospital’s rule, we get

$\mathop {\lim }\limits_{x \to \pi /2} \,\frac{{\cos x - \sin 2x}}{{ - \sin x}} = 0$.

$\mathop {\lim }\limits_{x \to 0} \,\frac{{\cos x - \frac{1}{{1 - x}}}}{{2x}} = \mathop {\lim }\limits_{x \to 0} \,\frac{{ - \sin x - \frac{1}{{{{(1 - x)}^2}}}}}{2} = - \frac{1}{2}$.

Aliter : $\mathop {\lim }\limits_{x \to 0} \,\frac{{\sin x + \log \,(1 - x)}}{{{x^2}}}$

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

$\left( {\because \sin x = x - \frac{{{x^3}}}{{3\,!}} + \frac{{{x^5}}}{{5\,!}} - ..} \right.$ 

and  $\left. {\log \,(1 - x) = - x - \frac{{{x^2}}}{2} - \frac{{{x^3}}}{3} - ..} \right)$

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

$= \mathop {\lim }\limits_{x \to \infty } \,{\left\{ {{{\left( {1 + \frac{1}{{a + bx}}} \right)}^{a + bx}}} \right\}^{\frac{{c + dx}}{{a + bx}}}} = {e^{d/b}}$

$\left\{ {\because \,\,\mathop {\lim }\limits_{x \to \infty } \,{{\left( {1 + \frac{1}{{a + bx}}} \right)}^{a + bx}} = e} \right.$  and 

$\left. {\mathop {\lim }\limits_{x \to \infty } \frac{{c + dx}}{{a + bx}} = \frac{d}{b}} \right\}$.

$ = \mathop {\lim }\limits_{n \to \infty } \,5\,{\left[ {{{\left\{ {1 + {{\left( {\frac{4}{5}} \right)}^n}} \right\}}^{{{(5/4)}^n}}}} \right]^{(1/n)\,.\,{{(4/5)}^n}}} = 5\,.\,{e^0} = 5$.

$\left( {\because \,\,{{\left( {\frac{4}{5}} \right)}^n} \to 0\,{\text{as }}\,n \to \infty } \right)$

$= \mathop {\lim }\limits_{x \to \infty } \,\sqrt {\left( {\frac{{1 + \frac{{\sin x}}{x}}}{{1 - \frac{{\cos x}}{x}}}} \right)} = \mathop {\lim }\limits_{x \to \infty } \sqrt 1 = 1$

$[\,\because \,\,\mathop {\lim }\limits_{x \to \infty } \,\frac{{\sin x}}{x}$ and $\mathop {\lim }\limits_{x \to \infty } \,\frac{{\cos x}}{x}$ both are equal to  $0$ ]

$ = y\mathop {\lim }\limits_{n \to \infty } \,\,{\left[ {{{\left( {1 + {{\left( {\frac{x}{y}} \right)}^n}} \right)}^{{{\left( {\frac{y}{x}} \right)}^n}.}}} \right]^{\frac{1}{n}.{{\left( {\frac{x}{y}} \right)}^n}}}$

$ = y{e^0} = y$, 

$\left[ {\because \,\,\frac{x}{y} < 1\, \Rightarrow \,\,{{\left( {\frac{x}{y}} \right)}^n} \to 0\,\,{\text{as}}\,\,n \to \infty } \right]$.

$\therefore \,\,\,\mathop {\lim }\limits_{x \to - 1} \,\frac{{\sqrt \pi - \sqrt {{{\cos }^{ - 1}}x} }}{{\sqrt {x + 1} }} = \mathop {\lim }\limits_{y \to \pi } \,\frac{{\sqrt \pi - \sqrt y }}{{\sqrt {1 + \cos y} }}$

$ = \mathop {\lim }\limits_{y \to \pi } \,\frac{{\sqrt \pi - \sqrt y }}{{\sqrt 2 \,\cos \,(y/2)}}\, $

$= \mathop {\lim }\limits_{y \to \pi } \,\,\frac{{\sqrt \pi - \sqrt y }}{{\sqrt 2 \,\sin \,\left( {\frac{\pi }{2} - \frac{y}{2}} \right)}}\frac{{\left( {\frac{\pi }{2} - \frac{y}{2}} \right)}}{{\left( {\frac{\pi }{2} - \frac{y}{2}} \right)}}$

$ = \mathop {\lim }\limits_{y \to \pi } \,\frac{1}{{\frac{{\sqrt 2 }}{2}(\sqrt \pi + \sqrt y )}}.\frac{1}{{\frac{{\sin \,\left( {\frac{\pi }{2} - \frac{y}{2}} \right)}}{{\left( {\frac{\pi }{2} - \frac{y}{2}} \right)}}}} = \frac{1}{{\sqrt {2\pi } }}.$

$ = \mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt {x + \sqrt x } }}{{\sqrt {x + \sqrt {x + \sqrt x } } + \sqrt x }} = \mathop {\lim }\limits_{x \to \infty } \,\frac{{\sqrt {1 + {x^{ - 1/2}}} }}{{\sqrt {1 + \sqrt {{x^{ - 1}} + {x^{ - 3/2}}} } + 1}} = \frac{1}{2}$.

$ = \frac{{{x^2} - 8x + 15}}{{{x^2} + x - 12}} = \frac{{(x - 3)\,(x - 5)}}{{(x - 3)\,(x + 4)}}$

$\therefore \,\,\mathop {\lim }\limits_{x \to 3} \,[f(x) + g(x) + h(x)] = \mathop {\lim }\limits_{x \to 3} \,\,\frac{{(x - 3)\,\,(x - 5)}}{{(x - 3)\,\,(x + 4)}} = - \frac{2}{7}$.

$ = \mathop {\lim }\limits_{x \to 2} \,\,\frac{{1 + \sqrt {2 + x} - 3}}{{(\sqrt {1 + \sqrt {2 + x} + \sqrt 3 )\,\,(x - 2)} }}$

$ = \mathop {\lim }\limits_{x \to 2} \,\,\frac{{\sqrt {2 + x} - 2}}{{(\sqrt {1 + \sqrt {2 + x} + \sqrt 3 )\,\,(x - 2)} }}$

$ = \mathop {\lim }\limits_{x \to 2} \,\,\frac{{(x - 2)}}{{(\sqrt {1 + \sqrt {2 + x} } + \sqrt 3 )\,\,(\sqrt {2 + x} + 2)\,\,(x - 2)}}$

$ = \frac{1}{{(2\sqrt 3 )\,4}} = \frac{1}{{8\sqrt 3 }}.$

$ = \frac{1}{{\sqrt 2 }}\,\mathop {\lim }\limits_{x \to 0} \,\left( {\frac{{\frac{{\sin \,({x^2}/2)}}{{{x^2}/2}}}}{{{{\left( {\frac{{\sin \,(x/2)}}{{x/2}}} \right)}^2}}}} \right).\frac{{{x^2}/2}}{{{x^2}/4}} = \sqrt 2 $. 

$= \mathop {\lim }\limits_{x \to \infty } \,\,\frac{{{x^2}\left( {2 - \frac{3}{3}\,} \right)\,\left( {3 - \frac{4}{x}} \right)}}{{{x^2}\left( {4 - \frac{5}{x}} \right)\,\left( {5 - \frac{6}{x}} \right)}} = \frac{6}{{20}} = \frac{3}{{10}}$

$ = \mathop {\lim }\limits_{t \to 0} \,\,\frac{{\sin \,({e^t} - 1)}}{{\log \,(1 + t)}}$,          $\{$Putting $x = 2 + t\} $

$ = \mathop {\lim }\limits_{t \to 0} \,\,\frac{{\sin \,({e^t} - 1)}}{{{e^t} - 1}}.\frac{{{e^t} - 1}}{t}.\frac{t}{{\log \,(1 + t)}}$

$ = \mathop {\lim }\limits_{t \to 0} \,\,\frac{{\sin \,({e^t} - 1)}}{{{e^t} - 1}}.\left( {\frac{1}{{1\,\,!}} + \frac{t}{{2\,\,!}} + ...} \right) \times \left[ {\frac{1}{{\left( {1 - \frac{1}{2}t + \frac{1}{3}{t^2} - ...} \right)}}} \right]$

$ = 1\,\,.\,\,1\,\,.\,\,1 = 1,\,\,\,\,(\because \,\,{\text{As}}\,\,t \to 0,\,\,{e^t} - 1 \to 0).$

$ = \mathop {\lim }\limits_{x \to \infty } \,\frac{4}{{\left( {\sqrt {1 + \frac{8}{x} + \frac{3}{{{x^2}}}} + \sqrt {1 + \frac{4}{x} + \frac{3}{{{x^2}}}} } \right)}} = 2$.

$\therefore$ $\mathop {\lim }\limits_{x \to 5} \frac{{{x^k} - {5^k}}}{{x - 5}} = k{(5)^{k - 1}}$; 

But given, $\mathop {\lim }\limits_{x \to 5} \,\frac{{{x^k} - {5^k}}}{{x - 5}} = 500$,

$\therefore$ $k{(5)^{k - 1}} = 500$; $k\,{(5)^{k - 1}} = 4\,{(5)^{4 - 1}},$

$\therefore \,\,\,k = 4$.

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

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

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

$\therefore \,\,\,\mathop {\lim }\limits_{x \to 0} \,\,f(x) = 0$,    $\left( {\because \mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} f(x)} \right)$ .

Applying $ L-$ Hospital's rule,

$y = \mathop {\lim }\limits_{x \to 0} \,\,\frac{{{e^x} + x\,{e^x} - \frac{1}{{1 + x}}}}{{2x}}$,   $\left( {\frac{0}{0}\,{\rm{form}}} \right)$

$y = \mathop {\lim }\limits_{x \to 0} \,\,\frac{1}{2}\,\left[ {{e^x} + {e^x} + x\,{e^x} + \frac{1}{{{{(1 + x)}^2}}}} \right]$

$y = \mathop {\lim }\limits_{x \to 0} \,\,\frac{1}{2}\,[1 + 1 + 0 + 1] = \frac{3}{2}$.

$ = \mathop {\lim }\limits_{h \to 0} \,\,\frac{{\sqrt {4\,{{( - 1/h)}^2} + 5\,( - 1/h) + 8} }}{{4\,( - 1/h) + 5}}$

$ = \mathop {\lim }\limits_{h \to 0} \,\,\frac{{(1/h)\sqrt {4\, - 5h + 8{h^2}} }}{{(1/h)\,( - \,4 + 5h)}} = \frac{{\sqrt 4 }}{{ - 4}} = - \frac{1}{2}$.

$ \Rightarrow \,\,y = {e^{1/m}},\,\,\,\left( {\because \mathop {\lim }\limits_{x \to \,\infty } \,{{\left( {1 + \frac{1}{x}} \right)}^x} = e} \right)$

$ = \mathop {\lim }\limits_{h \to 0} \,\,(3 - 3h) = 3 - 3\,.\,0 = 3$

$R.H.L.$$ = \mathop {\lim }\limits_{x \to 1 + 0} f(x) = \mathop {\lim \,\,}\limits_{h \to 0} \,f\,(1 + h) = \mathop {\lim }\limits_{h \to 0} \,\,[5 - 3(1 + h)]$

$ = \mathop {\lim }\limits_{h \to 0} \,\,(2 - 3h) = 2 - 3\,.\,0 = 2$

Hence $\mathop {\lim }\limits_{x \to 1} \,\,f(x)$ does not exist.

Applying $L-$ Hospital's rule, we get

$\mathop {\lim }\limits_{x \to 2} \,\,\frac{{3{x^2}}}{{2x}} = \mathop {\lim }\limits_{x \to 2} \,\,\frac{{3 \times 2 \times 2}}{{2 \times 2}} = 3$.

Applying  $ L-$ Hospital's rule, we get

$y = \mathop {\lim }\limits_{x \to 3} \,\,3{x^2} - 2x = (27 - 6) = 21$.

$ = \mathop {\lim }\limits_{x \to 0} \,\frac{{\frac{1}{{1 + {x^2}}}}}{1} = \mathop {\lim }\limits_{x \to 0} \,\,\frac{1}{{1 + {x^2}}} = \frac{1}{{1 + 0}} = 1$.

==> $y = \mathop {\lim }\limits_{x \to 4} \frac{{({x^{1/2}} - 2)(x + 4 + 2\sqrt x )}}{{(\sqrt x - 2)(\sqrt x + 2)}}$

==> $y = \mathop {\lim }\limits_{x \to 4} \frac{{(x + 4 + 2\sqrt x )}}{{(\sqrt x + 2)}}$$ = \frac{{4 + 4 + 2\sqrt 4 }}{{\sqrt 4 + 2}}$$ = \frac{{12}}{4} = 3$.

Trick : Applying  $ L-$ Hospital’s rule, we get

$\mathop {\lim }\limits_{x \to 4} \frac{{\frac{3}{2}{x^{1/2}}}}{1}$$ = \frac{3}{2}{(4)^{1/2}} = 3.$

Applying $ L-$ Hospital’s rule,

$\mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{{\sqrt {1 - {x^2}} }} - \frac{1}{{1 + {x^2}}}}}{{3{x^2}}}$,      $\left( {\frac{0}{0}} \,\,form \,\, \right)$

$ = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{ - 1}}{2} \times \frac{{ - 2x}}{{{{(1 - {x^2})}^{3/2}}}} + \frac{{2x}}{{{{(1 + {x^2})}^2}}}}}{{6x}}$

$ = \mathop {\lim }\limits_{x \to 0} \frac{1}{6}\left[ {\frac{1}{{{{(1 - {x^2})}^{3/2}}}} + \frac{2}{{{{(1 + {x^2})}^2}}}} \right]\, = \,\frac{1}{2}$.

$ = \mathop {{\rm{lim}}}\limits_{x \to \infty } \,{\left\{ {{{\left( {1 + \frac{{a - b}}{{x + b}}} \right)}^{\frac{{x + b}}{{a - b}}}}} \right\}^{a - b}} = {e^{a - b}}$.

$ = \mathop {{\rm{lim}}}\limits_{x \to \pi /2} {a^{\cos x}}\left( {\frac{{{a^{\cot x - \cos x}} - 1}}{{\cot x - \cos x}}} \right)$

$ = {a^{\cos (\pi /2)}}\mathop {{\rm{lim}}}\limits_{x \to \pi /2} \left( {\frac{{{a^{\cot x - \cos x}} - 1}}{{\cot x - \cos x}}} \right)$$ = 1.\log a = \log a$.

$ = {x^2}\sin x - {x^3}\cos x - {x^3}\tan x + 2{x^2}\cos x - x\sin x$

Hence $\mathop {\lim }\limits_{x \to 0} \frac{{f(x)}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \,\left( {\sin x - x\cos x - x\tan x + 2\cos x - \left. {\frac{{\sin x}}{x}} \right)} \right.$

$ = 0 - 0 - 0 + 2 - 1 = 1$.

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

$g(x) = {\cos ^{ - 1}}\left\{ {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right\}$

Put $x = \tan \theta $ in both equations

$f(\theta ) = {\cot ^{ - 1}}\left\{ {\frac{{3\tan \theta - {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }}} \right\}$$ = {\cot ^{ - 1}}\left\{ {\tan 3\theta } \right\}$

$f(\theta ) = {\cot ^{ - 1}}\cot \left( {\frac{\pi }{2} - 3\theta } \right) = \frac{\pi }{2} - 3\theta \Rightarrow f'(\theta ) = - 3$ .….(i)

and $g(\theta ) = {\cos ^{ - 1}}\left\{ {\frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right\}$$ = {\cos ^{ - 1}}(\cos 2\theta ) = 2\theta $

$ \Rightarrow g'(\theta ) = 2$ …..(ii)

Now $\mathop {\lim }\limits_{x \to a} \left( {\frac{{f(x) - f(a)}}{{g(x) - g(a)}}} \right) = \mathop {\lim }\limits_{x \to a} \left( {\frac{{f(x) - f(a)}}{{x - a}}} \right)\frac{1}{{\mathop {\lim }\limits_{x \to a} \left( {\frac{{g(x) - g(a)}}{{x - a}}} \right)}}$

$ = f'(x).\frac{1}{{g'(x)}} = - 3 \times \frac{1}{2} = - \frac{3}{2}$.

Using $ L-$ Hospital’s rule

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

==> $y = \frac{1}{{ - 4 + 2}} = - \frac{1}{2}$.

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

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

$ = \mathop {{\rm{lim}}}\limits_{n \to \infty } \,{({4^n})^{\frac{1}{n}}}{\left[ {\frac{{{3^n}}}{{{4^n}}} + 1} \right]^{\frac{1}{n}}}$

$ = \mathop {{\rm{lim}}}\limits_{n \to \infty } 4\,{\left[ {1 + \frac{1}{{{{\left( {\frac{4}{3}} \right)}^n}}}} \right]^{1/n}}$

$ = 4\mathop {{\rm{lim}}}\limits_{n \to \infty } \,{\left[ {1 + \frac{1}{{{{\left( {\frac{4}{3}} \right)}^n}}}} \right]^{1/n}}$

$ = 4{\left[ {1 + \frac{1}{\infty }} \right]^0} = 4 \times {(1)^0}$ $ = 4 \times 1 = 4$.