Maths M.C.Q (1 Marks)

$ = \mathop {\lim }\limits_{x \to \frac{\pi }{2}} \frac{{\frac{1}{{\sin x}}\cos x}}{{ - {\rm{cose}}{{\rm{c}}^{\rm{2}}}x}} = 0$     (Applying $ L-$ Hospital’s rule)

and $\mathop {\lim }\limits_{x \to 0 + } f(x) = \mathop {\lim }\limits_{h \to 0} \,\,\frac{h}{{h + {h^2}}} = \mathop {\lim }\limits_{h \to 0} \,\frac{1}{{1 + h}} = 1$

Hence limit does not exist.

$\mathop {\lim }\limits_{x \to a} \frac{{f(x)}}{{g(x)}} = \mathop {\lim }\limits_{x \to a} \frac{{f'(x)}}{{g'(x)}}$.

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

Aliter : Apply  $ L$-Hospital’s rule,

$\mathop {\lim }\limits_{x \to 0} \,\frac{{\cos ax - \cos bx}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \,\,\frac{{ - \,a\sin ax + b\sin bx}}{{2x}}$

$ = \mathop {\lim }\limits_{x \to 0} \,\frac{{ - \,{a^2}\cos ax + {b^2}\cos bx}}{2} = \frac{{{b^2} - {a^2}}}{2}.$

Aliter : Apply  $ L$-Hospital’s rule.

$\Rightarrow \,\,\frac{{2{a^9}}}{{2a}} = 9\,\, \Rightarrow \,\,{a^8} = 9$

$ \Rightarrow \,\,\,\,a = {9^{1/8}}$

as ${e^{ - 1/x}} \to 0$ when $x \to {0^ + }$

Aliter : $\mathop {\lim }\limits_{x \to 0} \,\,\frac{{\frac{{2\,\,\sin 2x}}{{2x}} + \frac{{6\,\,\sin 6x}}{{6x}}}}{{\frac{{5\,\,\sin 5x}}{{5x}} - \frac{{3\,\,\sin 3x}}{{3x}}}} = \frac{{2 + 6}}{{5 - 3}} = 4.$

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

$ = {\log _e}[{e^{\mathop {\lim }\limits_{x \to 0} \,x(\sin x - 1)}}] = {\log _e}1.$

$ = \log \,\,a - \log \,\,b = \log \,(a/b)$.

$ = \mathop {\lim }\limits_{x \to \infty } \frac{{[{e^{{{\log }_e}{a^{1/x}}}} - 1]}}{{1/x}} = {\log _e}a = - {\log _e}\frac{1}{a}.$

$\mathop {\lim }\limits_{x \to \alpha } \,\,\frac{{\cos x}}{1} = \cos \alpha $, (Apply  $L-$Hospital's rule)

$= \frac{{{a^2} - {b^2}}}{{{c^2} - {d^2}}}.$

Expand $sin\ x$, then

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

$= \frac{{ - 1}}{{3\,!}} = \frac{{ - 1}}{6}$.

Aliter : Apply  $L-$ Hospital’s rule.

Aliter : Apply  $ L-$ Hospital’s rule.

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

(By $L-$ Hospital’s rule)

$ = \mathop {\lim }\limits_{x \to 0} \,\,\frac{{ - \cos x}}{{3\cos x - x\sin x}} = - \frac{1}{3}$, 

(Again by $L-$ Hospital’s rule)

$ = - \frac{1}{3}$

Aliter : Apply  $ L-$ Hospital’s rule,

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

Aliter : Apply  $L-$ Hospital’s rule.

Aliter : Apply $ L-$ Hospital’s rule.

So $\mathop {\lim }\limits_{y \to 0} \frac{{\sqrt {1 + \sin y} - \sqrt {1 - \sin y} }}{y}$

$(\because \,\,\,x \to 0 \Rightarrow y \to 0)$

$( $ Now multiply it by $\frac{\sqrt{1+\sin y}+\sqrt{1-\sin y}}{\sqrt{1+\sin y}+\sqrt{1-\sin y}}$ and solve $) $

$= 1$

Aliter : Apply  $ L-$ Hospital’s rule.

Aliter : Apply  $ L-$ Hospital’s rule.

Aliter : $\mathop {\lim }\limits_{\theta \to 0} \,\frac{{\sin 3\theta - \sin \theta }}{{\sin \theta }} $

$= \mathop {\lim }\limits_{\theta \to 0} \,\frac{{\sin 3\theta }}{{\sin \theta }} - \mathop {\lim }\limits_{\theta \to 0} \,\frac{{\sin \theta }}{{\sin \theta }}$

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

You may also apply $L-$ Hospital rule.

Aliter : Apply $L-$ Hospital’s rule.

$ = \frac{{5 - 2}}{{3 + 1}} = \frac{3}{4}$

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

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

You may also apply $L-$ Hospital rule.

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

Hence limit does not exist.

$= \mathop {\lim }\limits_{x \to a} {\sin ^3}x = {\sin ^3}a$.

$ = \mathop {\lim }\limits_{h \to 0} \,\frac{{\frac{4}{{\sqrt 3 }}\,\left[ {\frac{{\sqrt 3 }}{2}\sin \,\left( {\frac{\pi }{6} + h} \right) - \frac{1}{2}\cos \,\left( {\frac{\pi }{6} + h} \right)} \right]}}{{h\,(\sqrt 3 \cos \,h - \sin \,h)}}$

$ = \mathop {\lim }\limits_{h \to 0} \frac{4}{{\sqrt 3 }}.\frac{{\sin \,h}}{h}.\frac{1}{{(\sqrt 3 \,\cos \,h - \sin \,h)}} = \frac{4}{3}$.

$\therefore \,\,\mathop {\lim }\limits_{y \to 0} \,\log y = \mathop {\lim }\limits_{x \to 0} x\log x$

$ = 0 = \log 1\,\, \Rightarrow \,\,\mathop {\lim }\limits_{x \to 0} {x^x} = 1$.

Aliter : Apply  $ L-$ Hospital’s rule two times.

and as $x \to \,(\pi /2),\,\,\theta \to 0$

Now solve yourself.

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

Now expanding ${e^{{x^2}}}$ and $\cos x,$ we get

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

Aliter : Apply $L-$ Hospital’s rule,

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

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

$ = \mathop {\lim }\limits_{x \to 0} \,\frac{{{e^{\alpha x}} - 1}}{{x}} - \mathop {\lim }\limits_{x \to 0} \,\frac{{{e^{\beta x}} - 1}}{{ x}}$

$ = \alpha \mathop {\lim }\limits_{x \to 0} \,\frac{{{e^{\alpha x}} - 1}}{{\alpha x}} - \beta \mathop {\lim }\limits_{x \to 0} \,\frac{{{e^{\beta x}} - 1}}{{\beta x}}$

$ = \alpha .1 - \beta .1 = \alpha - \beta .$

$= \mathop {\lim }\limits_{x \to a} \,\frac{{ - 1}}{{ax}} = \frac{{ - 1}}{{{a^2}}}$.

Or $\mathop {\lim }\limits_{x \to a} \,\,\frac{-1/{{x}^{2}}-0}{1-0}$

(By Appling $L-$ Hospital’s rule) 

$-\frac{1}{{{a}^{2}}}$

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