Maths M.C.Q (1 Marks)

$ = \mathop {\lim }\limits_{x \to \infty } \,\frac{{ax}}{{\,\sqrt {{a^2}{x^2} + ax + 1} + \sqrt {{a^2}{x^2} + 1} }}$

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

$ = \mathop {\lim }\limits_{x \to 0} \,{e^x}\,.\mathop {\lim }\limits_{x \to 0} \,\frac{{{e^{\tan x - x}} - 1}}{{\tan x - x}} = {e^0} \times 1 = 1$.

$= \mathop {\lim }\limits_{x \to \infty } \,\,\sqrt {\frac{{1 - \frac{{\sin x}}{x}}}{{1 + \frac{{{{\cos }^2}x}}{x}}}} $

$ = \sqrt {\frac{{1 - 0}}{{1 + 0}}} = 1$, 

$\left( {\because \,\,\,\frac{{\sin x}}{x} \to 0,\frac{{{{\cos }^2}x}}{x}\, \to 0\,\,{\text{as }}x \to \infty } \right)$.

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

$ = \mathop {\lim }\limits_{x \to \pi /2} \,\,\frac{{2x\,\,\sin x - \pi }}{{2\,\cos x}}$,   $\left[ {{\rm{form}} \,\, \frac{0}{0}} \right]$

$ = \mathop {\lim }\limits_{x \to \pi /2} \,\,\frac{{[2\,\sin x + 2x\cos x]}}{{ - 2\,\sin x}} = - 1$,   (By $L  -$ Hospital’s rule).

$ = - 2\,\sin a\,.\,\frac{{(1 - \cos x)}}{{{x^2}}}\,.\,\left( {\frac{x}{{\sin x}}} \right)$

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

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

$=$ (finite number lies between $-1$ to $1$)

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

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

$=$ (finite number lies between $0$ to $1$)

$ \because \,\,\,\,\mathop {\lim }\limits_{x \to {0^ - }} f(x) \ne \mathop {\lim }\limits_{x \to {0^ + }} f(x)$

$\therefore$ $\mathop {\lim }\limits_{x \to 0} \sin \left( {\frac{1}{x}} \right)$ does not exist.

$ = \mathop {{\rm{lim}}}\limits_{x \to 0} \frac{1}{e} = {e^{ - 1}}$.

Applying $ L-$ Hospital’s rule, we have

$\mathop {\lim }\limits_{x \to 0} \,\frac{{\cos x - x\sin x - \frac{1}{{x + 1}}}}{{2x}}$,          $\left( {\frac{0}{0}{\rm{form}}} \right)$

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

$= \mathop {\lim }\limits_{a \to 0} \frac{{ - (1 - \cos a)}}{{(1 - {{\cos }^2}a)(\cos a)}}$

$ = \mathop {\lim }\limits_{a \to 0} \left[ { - \frac{1}{{(1 + \cos a)\cos a}}} \right] = - \frac{1}{{(1 + 1)1}} = \frac{{ - 1}}{2}$

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

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

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

Again applying $ L-$ Hospital’s rule,

$\mathop {{\rm{lim}}}\limits_{x \to 1} \frac{{ - 1}}{{4x - 2}} = - \frac{1}{2}$.

$ = {\log _e}a \times \frac{1}{{{{\log }_e}b}} = \frac{{\log a}}{{\log b}}$.

$ = \mathop {{\rm{lim}}}\limits_{x \to 2} \,\frac{1}{{{3^{x/2}} + 3}} = \frac{1}{6}$.

$ = \mathop {{\rm{lim}}}\limits_{x \to 0} \frac{{2{{\sin }^2}x\,\sin 5x}}{{{x^2}\sin 3x}}$

$ = \mathop {{\rm{lim}}}\limits_{x \to 0} \,\left( {\frac{{2{{\sin }^2}x}}{{{x^2}}}} \right)\frac{{\left( {\frac{{\sin 5x}}{x}} \right)}}{{\left( {\frac{{\sin 3x}}{x}} \right)}}$

$ = \mathop {{\rm{lim}}}\limits_{x \to 0} 2\,{\left( {\frac{{\sin x}}{x}} \right)^2} \times \frac{{5\mathop {{\rm{lim}}}\limits_{x \to 0} \left( {\frac{{\sin 5x}}{{5x}}} \right)}}{{3\mathop {{\rm{lim}}}\limits_{x \to 0} \left( {\frac{{\sin 3x}}{{3x}}} \right)}}$

$ = \frac{{2 \times 5}}{3} = \frac{{10}}{3}$.

$\mathop {\lim }\limits_{x \to 0} \frac{{\ln (\cos x)}}{{{x^2}}}$$ = \mathop {\lim }\limits_{x \to 0} \frac{{ - \tan x}}{{2x}}$

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

$ = \mathop {{\rm{lim}}}\limits_{x \to 0} \,\left[ {\frac{{(\sqrt {a + x} - \sqrt {a - x} )(\sqrt {a + x} + \sqrt {a - x} )}}{{x(\sqrt {a + x} + \sqrt {a - x} )}}} \right]$

$ = \mathop {{\rm{lim}}}\limits_{x \to 0} \,\left[ {\frac{{2x}}{{x(\sqrt {a + x} + \sqrt {a - x} )}}} \right] = \frac{2}{{\sqrt a + \sqrt a }} = \frac{1}{{\sqrt a }}$.

Applying $ L-$ Hospital’s rule,

$\mathop {{\rm{lim}}}\limits_{\alpha \to \beta } \frac{{2\sin \,\alpha \,\,\cos \alpha }}{{2\alpha }} = \mathop {{\rm{lim}}}\limits_{\alpha \to \beta } \frac{{\sin \,\,2\alpha }}{{2\alpha }} = \frac{{\sin \,\,2\beta }}{{2\beta }}$.

[Using  $ L-$ Hospital’s rule]

$ = \mathop {\lim }\limits_{x \to 1} \frac{{ - 1}}{2}{\cos ^3}\pi \,x$$ = \frac{1}{2}$.

$\therefore$ $\mathop {\lim }\limits_{x \to 3} \,\,[x]$ does not exist.

Using $ L-$ Hospital’s rule, 

$\mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{{1 + x}}}}{{{3^x}{{\log }_e}3}} = \frac{1}{{{{\log }_e}3}} = {\log _3}e$.

$\therefore$  Limit doesn’t exist.

Using $L-$ Hospital’s rule,

$y = \mathop {\lim }\limits_{x \to 0} \frac{{{4^x}\log 4 - {9^x}\log 9}}{{({4^x} + {9^x}) + x({4^x}\log 4 + {9^x}\log 9)}}$

==> $y = \frac{{\log 4 - \log 9}}{2}$

==> $y = \frac{{\log {{\left( {\frac{2}{3}} \right)}^2}}}{2} = \log \frac{2}{3}$.

$ = \mathop {\lim }\limits_{x \to 0} \left[ {\frac{{{a^x} - 1}}{x} - \frac{{{b^x} - 1}}{x}} \right]\frac{x}{{{e^x} - 1}}$

$ = ({\log _e}a - {\log _e}b).\frac{1}{{{{\log }_e}e}}$$ = {\log _e}\left( {\frac{a}{b}} \right)$

Trick : Apply  $ L-$ Hospital’s rule.

$\mathop {\lim }\limits_{x \to 7} \frac{{2 - \sqrt {x - 3} }}{{{x^2} - 49}}$ 

$= \mathop {\lim }\limits_{x \to 7} \frac{{0 - \frac{1}{{2\sqrt {x - 3} }}}}{{2x}}$

$ = \mathop {\lim }\limits_{x \to 7} \frac{{ - 1}}{{4x\sqrt {x - 3} }} = \frac{{ - 1}}{{4.7\sqrt {7 - 3} }} = \frac{{ - 1}}{{56}}$.

==> $y = \mathop {\lim }\limits_{x \to 0} \frac{{\left[ {1 + \frac{x}{{1!}} + \frac{{{x^2}}}{{2!}} + ....} \right] - \left[ {1 - \frac{x}{{1!}} + \frac{{{x^2}}}{{2!}} - ....} \right]}}{{\sin x}}$

==> $y = \mathop {\lim }\limits_{x \to 0} \frac{{2\,\left[ {\frac{x}{{1!}} + \frac{{{x^3}}}{{3!}} + \frac{{{x^5}}}{{5!}} + .............} \right]}}{{\sin x}}$

==> $y = \mathop {\lim }\limits_{x \to 0} \frac{{2\,\left[ {1 + \frac{{{x^2}}}{{3!}} + \frac{{{x^4}}}{{4!}} + ...........} \right]}}{{\frac{{\sin x}}{x}}}$

==> $y = \frac{{\mathop {\lim }\limits_{x \to 0} 2\,\left[ {1 + \frac{{{x^2}}}{{2!}} + .......} \right]}}{{\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x}}}$

==> $y = \frac{2}{1} = 2$

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

$\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - {e^{ - x}}}}{{\sin x}}$

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

$\mathop {\lim }\limits_{x \to \pi /6} \frac{{3\cos x + \sqrt 3 \sin x}}{6} = \frac{{3.\frac{{\sqrt 3 }}{2} + \sqrt 3 .\frac{1}{2}}}{6} = \frac{1}{{\sqrt 3 }}$.

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

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

$ = \mathop {\lim }\limits_{x \to 0} 2{\log _e}e = 2$

$\left\{ { \because \mathop {\lim }\limits_{x \to 0} {{(1 + x)}^{\frac{1}{x}}} = {{\log }_e}e = 1} \right\}$

Trick : Using  $L$  Hospital’s rule.

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

$ = \mathop {\lim }\limits_{x \to \infty } {e^{\frac{{ - 2(x + 1)}}{{3x + 2}}}} = {e^{ - 2/3}}$,      $\left\{ \because \underset{x\to \infty }{\mathop{\lim }}\,\frac{-2(x+1)}{3x+2}=\frac{-2}{3} \right\}$

$ = \mathop {\lim }\limits_{x \to \infty } \left[ {\frac{1}{x}\frac{{\left( {1 + \frac{1}{x}} \right)\,\left( {3 + \frac{4}{x}} \right)}}{{\left( {1 - \frac{8}{x}} \right)}}} \right] = 0$.

$\mathop {\lim }\limits_{n \to \infty } \,\left[ {1 - \frac{1}{{{2^n}}}} \right] = 1 - 0 = 1$

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

$ = \frac{1}{2}\,\,\mathop {\lim }\limits_{n \to \infty } \,\,\frac{{n + 1}}{n} = \frac{1}{2}\,\,\mathop {\lim }\limits_{n \to \infty } \,\,1 + \frac{1}{n} = \frac{1}{2}$

$ = \frac{{ - 2}}{{1 + 2}} = \frac{{ - 2}}{3}$.

$ = {\left. {f'(x)} \right|_{x = 0}}$ $ = {\left. {{{\tan }^4}x} \right|_{x = 0}} = 0$

$ = \mathop {\lim }\limits_{x \to 0} \,{[{\{ 1 + \tan x)^{\cot \,x}}\} ^{sec\,x}} \times \{ 1/{(1 + \sin x)^{\cos ec\,x}}\} ]$

$ = {e^{\sec \,\,0}}.\frac{1}{e} = e\,.\,\frac{1}{e} = 1.$

$ = \mathop {\lim }\limits_{t \to 0} \,\,\frac{{\frac{{\sin t}}{t} - 1}}{{t - 1}} = \frac{{1 - 1}}{{0 - 1}} = 0,$

which is given in $(a)$.

Aliter : $\mathop {\lim }\limits_{x \to \infty } \,\,\frac{{{x^2}\sin \frac{1}{x} - x}}{{1 - \,\,|x|}}$

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

$ = \mathop {\lim }\limits_{x \to \infty } \,\,\frac{{\left( {x - \frac{1}{{6x}} + .... - x} \right)}}{{1 - |x|}}$

$ = \mathop {\lim }\limits_{x \to \infty } \,\frac{{\frac{1}{{6x}} - {\rm{terms \,containing \,powers\, of\, }}\frac{1}{x}}}{{|x| - 1}} = 0.$

$\Rightarrow$ $\log y = \mathop {\lim }\limits_{x \to 0} \, \frac{2}{x} \log \left( {\frac{{{a^x} + {b^x} + {c^x}}}{3}} \right)$

$ = 2\,\mathop {\lim }\limits_{x \to 0} \,\frac{{\log \,({a^x} + {b^x} + {c^x}) - \log 3}}{x}$

Now applying $L-$ Hospital’s rule, we have

$\log y = \log \,{(abc)^{2/3}}\, \Rightarrow \,\,y = {(abc)^{2/3}}$

$ = \mathop {\lim }\limits_{x \to 0 + } \,\frac{{n\,{{(\log x)}^{n - 1} \frac{1}{x}}}}{{ - m{x^{ - m-1}}}}\,$    (By $L-$ Hospital's rule)

$ = \mathop {\lim }\limits_{x \to 0 + } \,\frac{{n\,{{(\log x)}^{n - 1}}}}{{ - m{x^{ - m}}}}\,$        $\left( {{\rm{Form}} \,\, \frac{\infty }{\infty }} \right)$

$ = \mathop {\lim }\limits_{x \to 0 + } \,\frac{{n\,(n - 1)\,{{(\log x)}^{(n - 2)}}\frac{1}{x}}}{{{{( - m)}^2}{x^{ - m - 1}}}}$        (By $L-$ Hospital's rule)

$ = \mathop {\lim }\limits_{x \to 0 + } \,\frac{{n\,(n - 1)\,{{(\log x)}^{n - 2}}}}{{{m^2}{x^{ - m}}}}\,$       $\left( {{\rm{Form}} \,\, \frac{\infty }{\infty }} \right)$
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$ = \mathop {\lim }\limits_{x \to 0 + } \,\frac{{n\,\,!}}{{{{( - m)}^n}{x^{ - m}}}} = 0$

(Differentiating ${N^r}$ and ${D^r}$   $n$ times).

$ = {e^{1 - \frac{x}{2} + \frac{{{x^2}}}{3}\, - ......}} = e\,{e^{ - \,\frac{x}{2} + \frac{{{x^2}}}{3}\, - \,.....}}$

$ = e\,\left[ {1 + \left( { - \frac{x}{2} + \frac{{{x^2}}}{3} - .....} \right) + \frac{1}{{2\,\,!}}\,{{\left( { - \frac{x}{2} + \frac{{{x^2}}}{3}\, - \,.....} \right)}^2} + ...} \right]$

$ = e\,\left[ {1 - \frac{x}{2} + \frac{{11}}{{24}}{x^2} - ....} \right]$

$\therefore \,\,\,\mathop {\lim }\limits_{x \to 0} \,\,\frac{{{{(1 + x)}^{1/x}} - e + \frac{{ex}}{2}}}{{{x^2}}} = \frac{{11e}}{{24}}$.

$ = {e^{\frac{1}{x}\,\left( {x\, - \,\frac{{{x^2}}}{2}\, + \,\frac{{{x^3}}}{3}\, - \,\frac{{{x^4}}}{4}\, + ....} \right)}}$$ = {e^{\left( {1\, - \,\frac{x}{2}\, + \,\frac{{{x^2}}}{3}\, - \,\frac{{{x^3}}}{4}\, + \,....} \right)}}$

$ = e.{e^{\left( {\, - \,\frac{x}{2}\, + \,\frac{{{x^2}}}{3}\, - \,\frac{{{x^3}}}{4} + ....} \right)}}$

$ = e\left[ {\frac{{\left( { - \frac{x}{2} + \frac{{{x^2}}}{3} - \frac{{{x^3}}}{4} + ...} \right)}}{{1!}} + \frac{{{{\left( { - \frac{x}{2} + \frac{{{x^2}}}{3} - \frac{{{x^3}}}{4} + ...} \right)}^2}}}{{2!}} + ...} \right]$

$ = \left[ {e - \frac{{ex}}{2} + \frac{{11e}}{{24}}{x^2} + ... + ...} \right]$

$\therefore$ $\mathop {\lim }\limits_{x \to 0} \frac{{{{(1 + x)}^{1/x}} - e}}{x}$$ = \mathop {\lim }\limits_{x \to 0} \,\left[ {\frac{{e - \frac{{ex}}{2} - \frac{{11e}}{{24}}{x^2} + ...e}}{x}} \right]$

==> $\mathop {\lim }\limits_{x \to 0} \,\left( { - \frac{e}{2} - \frac{{11e}}{{24}}x + ...} \right)$$ = - \frac{e}{2}$.

$ = \mathop {\lim }\limits_{m \to \infty } {\left[ {1 - \left( { - \cos \frac{x}{m} + 1} \right)} \right]^m}$

$ = \mathop {\lim }\limits_{m \to \infty } {\left[ {1 - 2{{\sin }^2}\frac{x}{{2m}}} \right]^m}$

$ = {e^{\mathop {\lim }\limits_{m \to \infty } - \left( {2{{\sin }^2}\frac{x}{{2m}}} \right)\,m}}$

$ = {e^{\mathop {\lim }\limits_{m \to \infty } - 2{{\left( {\frac{{\sin \frac{x}{{2m}}}}{{x/2m}}} \right)}^2}\left( {\frac{{{x^2}}}{{4{m^2}}}} \right)\,m}}$

$ = {e^{ - 2\mathop {\lim }\limits_{m \to \infty } \frac{{{x^2}}}{{4m}}}} = {e^0} = 1$.

$ = \mathop {\lim }\limits_{\theta \to 0} \left( {\frac{\theta }{{\sin \theta }}} \right)\frac{{2{{\sin }^2}\theta /2}}{{{{\sin }^2}\theta \cos \theta }}$

$ = \mathop {\lim }\limits_{\theta \to 0} \frac{{2{{\sin }^2}\theta /2}}{{(2\sin (\theta /2)\cos {{(\theta /2)}^2})}}\frac{1}{{\cos \theta }}$

$ = \mathop {\lim }\limits_{\theta \to 0} \frac{1}{2}\frac{1}{{{{\cos }^2}(\theta /2).\cos \theta }} = \frac{1}{2}$.