Maths M.C.Q (1 Marks)

$\lim _{x \rightarrow \infty}\left(\frac{x^2(1-a)+x(1-a-b)+(1-b)}{x+1}\right)=4$

Limit is finite

It exists when $\quad 1- a =0 \quad \Rightarrow a =1$

then

$\lim _{x \rightarrow \infty}\left(\frac{1-a-b+\frac{1-b}{x}}{1+\frac{1}{x}}\right)=4$

$\therefore \quad 1-a-b=4 \quad \Rightarrow \quad b=-4$

$\Rightarrow \sin ^2 \theta=\frac{1+b^2}{2 b}$

$\Rightarrow \sin ^2 \theta=1 \text { as } \frac{1+b^2}{2 b} \geq 1$

$\theta=\pm \frac{\pi}{2}$

$=\lim _{x \rightarrow 0} \frac{a-a\left(1-\frac{x^2}{a^2} \frac{1}{2}-\frac{x^2}{4}\right.}{x^4}$

$=\lim _{x \rightarrow 0} \frac{a-a\left(1-\frac{1 x^2}{2 a^2}+\frac{\frac{1}{2}\left(\frac{1}{2} 1\right)}{2}\left(\frac{-x^2}{a^2}\right)^2\right)^{-\frac{x^2}{4}}}{x^4}$

$=\lim _{x \rightarrow 0} \frac{a-a\left(1-\frac{1 x^2}{2 a^2} \frac{1 x^4}{8 a_4}\right)-\frac{x^2}{4}}{x^4}$

$=\lim _{ x \rightarrow 0} \frac{\frac{1 x ^2}{2 a }+\frac{1 x ^4}{8 x ^2} \frac{ x ^4}{4}}{ x ^4}$

$=\lim _{x \rightarrow 0} \frac{x^2\left(\frac{1}{2 a} \frac{1}{4}\right)+\frac{x^4}{8 a^3}}{x^4}$

Limit is finite, when $\frac{1}{2 a }=\frac{1}{4}$

$a =2$

$\therefore L =\frac{1}{8 a ^3}$

$\therefore L =\frac{1}{8 \times 4}$

$\therefore L =\frac{1}{64}$

$\therefore L =\frac{1}{64}, a =2$

When $x \rightarrow o$ then, for existence of $\frac{P(x)}{x^2} d=e=o$ and $\lim _{x \rightarrow 0}\left(1+\frac{P(x)}{x^2}\right)=2$

So $c=1$.

$P^{\prime}(1)=P^{\prime}(2)=0$

$4 a+3 b+2 c+d=0$

$4 a+3 b+2=0 \quad \ldots \ldots \ldots . \text { eq. } 1$

$4 a \cdot 8+3 b \cdot 4+2 c \cdot 2+d=0$

$32 a+12 b+4 c+d=0$

$32 a+12 b+4=0 \quad \ldots \ldots . \text { eq. }$

By solving eq. $1$ and eq. $2$ we get $a =\frac{1}{4} ; b =-1$

So, $P ( x )=\frac{1}{4} x ^4- x ^3+ x ^2$.

$P (2)=4-8+4=0$

$ = \mathop {\lim }\limits_{x \to 0} \frac{{2xf'({x^2}) - f'(x)}}{{f'(x)}}$,       (using $L'$  Hospital's rule)

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

as $f$ is strictly increasing.

==> $n((a - n)n - 1) = 0 \Rightarrow (a - n)n = 1 $

$\Rightarrow a = n + \frac{1}{n}$.

$ = \frac{{6 \times 2}}{{4 \times 1}} = 3$.

Limit $ = \mathop {\lim }\limits_{x \to 0} \frac{{2{{\sin }^2}\frac{x}{2}}}{{{2^2}{{(x/2)}^2}}}\frac{{{e^x} - \cos x}}{{{x^{n - 2}}}} = \frac{1}{2}\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - \cos x}}{{{x^{n - 2}}}}$

$(n \ne 1$ for then the limit $ = 0)$

$ = \frac{1}{2}\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} + \sin x}}{{(n - 2){x^{n - 3}}}}$.

So, if $n = 3,$ the limit is $\frac{1}{{2(n - 2)}}$ which is finite.

If $n = 4,$ the limit is infinite.

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

$ = \pi ( - 1).1.( - 1) = \pi $.

$\left( {\because \,\mathop {\lim }\limits_{x \to \infty } \frac{{ - 5x}}{{x + 2}} = \,\mathop {\lim }\limits_{x \to \infty } \frac{{ - 5}}{{1 + \frac{2}{x}}} =  - 5} \right)$.

$ = \mathop {\lim }\limits_{x \to 0} \,\frac{{x(\tan \,\,2x - 2\tan x)}}{{{{(2\,{{\sin }^2}x)}^2}}} = \mathop {\lim }\limits_{x \to 0} \,\,\frac{1}{4}\,\frac{{x\,(\tan 2x - 2\tan x)}}{{{{\sin }^4}x}}$

$ = \mathop {\lim }\limits_{x \to 0} \frac{1}{4}\frac{{x\left\{ {\left( {2x + \frac{1}{3}{{(2x)}^3} + \frac{2}{{15}}\,{{(2x)}^5} + ...} \right) - 2\left( {x + \frac{{{x^3}}}{3} + \frac{2}{{15}}{x^5} + ...} \right)} \right\}}}{{{x^4}\,{{\left( {1 - \frac{{{x^2}}}{{3\,\,!}} + \frac{{{x^4}}}{{5\,\,!}} + ....} \right)}^4}}}$

$ = \frac{1}{4}\,.\,\left( {\frac{8}{3} - \frac{2}{3}} \right) = \frac{2}{4} = \frac{1}{2}$.

$ \Rightarrow \;{\log _e}A\; = \;\mathop {\lim }\limits_{x \to 0} \frac{1}{{{x^2}}}\log \;\left( {1 + \frac{{2{x^2}}}{{1 + 3{x^2}}}} \right)$

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

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

$\therefore \;A\; = \;{e^2}$

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

$ = \mathop {\lim }\limits_{x \to \infty } \frac{{n\,\,!}}{{{e^x}}} = \frac{{n\,\,!}}{\infty } = 0$,

where $n$ is any whole number

$( \because n! $ is defined for all positive integers including zero).

$ = \mathop {\lim }\limits_{x \to \pi /4} \frac{{\sin x\,(1 + \cot x)}}{{(\sqrt 2 + \sec x)}} = \frac{{\frac{1}{{\sqrt 2 }}(2)}}{{\sqrt 2 + \sqrt 2 }} = \frac{1}{2}.$

Aliter : Apply $L-$ Hospital’s rule.

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

$\mathop {\lim }\limits_{h \to 0} \,\frac{{{{(a + h)}^2}\sin (a + h) - {a^2}\sin a}}{h}$

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

$ = 2a\,\,\sin a + {a^2}\cos \,\,a.$

Since $ - 1 \le \sin \frac{1}{x} \le 1\,\, \Rightarrow \,\, - |\,\,x\,\,|\,\, \le x\sin \frac{1}{x} \le \,\,|\,\,x\,\,|$

We know that $\mathop {\lim }\limits_{x \to 0} \,\,|\,\,x\,\,|\, = 0$ and $\mathop {\lim }\limits_{x \to 0} \,\,|\,\,x\,\,|\, = 0$

In this way $\mathop {\lim }\limits_{x \to 0} \,\,f(x) = 0.$

$\mathop {\lim }\limits_{x \to 9} \frac{{\frac{1}{{2\sqrt {f(x)} }} \cdot f'(x)}}{{\frac{1}{{2\sqrt x }}}} = \frac{{\frac{{f'(9)}}{{\sqrt {f(9)} }}}}{{\frac{1}{{\sqrt 9 }}}} = \frac{{\frac{4}{3}}}{{\frac{1}{3}}} = 4$

Therefore function is defined as $f(x) = \left\{ \begin{array}{l}\frac{{\sin \,[x]}}{{[x]}}\,\,( - 1 \le x < 0)\\\;\;\;\;\;0\;\;(0 \le x < 1)\end{array} \right.$

$\therefore$ Left hand limit $ = \mathop {\lim }\limits_{x \to 0 - } f(x) = \mathop {\lim }\limits_{x \to 0 - } \,\frac{{\sin \,[x]}}{{[x]}}$

$ = \frac{{\sin \,( - 1)}}{{ - 1}} = \sin {1^c}$

Right hand limit $= 0$. 

Hence limit doesn't exist.

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

$\left\{ \because \,\,\,\mathop {\lim }\limits_{x \to a} \,\,\frac{f(x)}{g(x)}=\mathop {\lim }\limits_{x \to a} \,\,\frac{{f}'(x)}{{g}'(x)} \right\}$

$ = 2\,\log 2 = \log 4.$

$ = \log \,\,2\,.\mathop {\lim }\limits_{x \to 0} \,\,\frac{{{x^2}}}{{2\,\,{{\sin }^2}\frac{x}{2}}} = (\log \,\,2)\,.\,2 = 2\log 2 = \log 4$.

Put $1 - x = y$ as $x \to 1,\,\,y \to 0$

Thus $\mathop {\lim }\limits_{y \to 0} \,\,y\tan \frac{{\pi \,(1 - y)}}{2} = \mathop {\lim }\limits_{y \to 0} \,\,\frac{2}{\pi }.\frac{{\left( {\frac{{\pi y}}{2}} \right)}}{{\tan \,\left( {\frac{{\pi y}}{2}} \right)}} $

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

$ = \mathop {\lim }\limits_{x \to a} \,\frac{{\sqrt {a + 2x} - \sqrt {3x} }}{{\sqrt {3a + x} - 2\sqrt x }} \times \frac{{\sqrt {a + 2x} + \sqrt {3x} }}{{\sqrt {a + 2x} + \sqrt {3x} }} \times \frac{{\sqrt {3a + x} + 2\sqrt x }}{{\sqrt {3a + x} + 2\sqrt x }}$

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

Aliter : Apply $L-$ Hospital’s rule.

$ = \mathop {\lim }\limits_{\alpha \to \pi /4} \,\left\{ {\frac{{\sqrt 2 \left( {\sin \alpha .\frac{1}{{\sqrt 2 }} - \cos \alpha .\frac{1}{{\sqrt 2 }}} \right)}}{{\left( {\alpha - \frac{\pi }{4}} \right)}}} \right\}$

$ = \sqrt 2 \,\mathop {\lim }\limits_{\alpha \to \pi /4} \,\frac{{\sin \,\left( {\alpha - \frac{\pi }{4}} \right)}}{{\left( {\alpha - \frac{\pi }{4}} \right)}} = \sqrt 2 \times 1 = \sqrt 2 $.

Aliter : Apply $L-$ Hospital’s rule,

$\mathop {\lim }\limits_{\alpha \to \pi /4} \,\frac{{\sin \alpha - \cos \alpha }}{{\alpha - (\pi /4)}} = \mathop {\lim }\limits_{\alpha \to \pi /4} \,\frac{{\cos \alpha + \sin \alpha }}{1} = \frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }} = \sqrt 2 $.

Aliter : Apply $L$- Hospital’s rule.

so that $x \to \infty \,\, \Rightarrow \,y \to 0$

$\therefore \,\mathop {\lim }\limits_{x \to \infty } \,\left( {\frac{{\sin x}}{x}} \right) = \mathop {\lim }\limits_{y \to 0} \,\left( {y.\sin \frac{1}{y}} \right)$ 

$= \mathop {\lim }\limits_{y \to 0} \,y \times \mathop {\lim }\limits_{y \to 0} \,\sin \frac{1}{y} = 0 \times ... = 0$

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

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

Aliter : Apply $L-$ Hospital‘s rule

$\mathop {\lim }\limits_{x \to 0} \,\,\frac{{\tan 2x - x}}{{3x - \sin x}} = \mathop {\lim }\limits_{x \to 0} \,\frac{{2{{\sec }^2}2x - 1}}{{3 - \cos x}} = \frac{{2 - 1}}{{3 - 1}} = \frac{1}{2}.$

$ = \mathop {\lim }\limits_{x \to 0} \,{\left( {\frac{x}{{\sin x}}} \right)^3} \times \mathop {\lim }\limits_{x \to 0} \,\cos x \times \mathop {\lim }\limits_{x \to 0} \,(1 + \cos x) = 2$

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

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

$ = \frac{2}{{2\sqrt {2a} }} = \frac{1}{{\sqrt {2a} }}$

Aliter : Apply $L$- Hospital’s rule

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

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

Aliter : $\mathop {\lim }\limits_{x \to 1 - } \,f(x) = \mathop {\lim }\limits_{h \to 0} \,\,(1 - h) = 1$

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

Hence limit of function is $1$.

Aliter : Apply $L-$ Hospital’s rule,

$\mathop {\lim }\limits_{x \to 1} \frac{{\log x}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\frac{1}{x}}}{1} = 1$

$\Rightarrow \,\,n.\,{2^{n - 1}} = 80\,\, \Rightarrow \,\,n = 5$.

$\Rightarrow \,\,k\,\,.\,\mathop {\lim }\limits_{x \to 0} \,\,\,\frac{x}{\sin x}=\frac{1}{k} \mathop {\lim }\limits_{x \to 0} \,\,\,\frac{kx}{\sin \,kx}$

$\Rightarrow \,\,k=\frac{1}{k}\,\,\Rightarrow \,\,k=\pm \,1$

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

Aliter : Apply $L-$ Hospital’s rule,

$\mathop {\lim }\limits_{x \to 0} \,\frac{{\log \cos x}}{x} = \mathop {\lim }\limits_{x \to 0} \,\frac{{ - \tan x}}{1} = 0.$

hence limit does not exist.

Aliter : Apply $L-$ Hospital rule,

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

$ = \mathop {\lim }\limits_{x \to 0} \,\left[ {{{\left\{ {\frac{{\sin {\textstyle{{mx} \over 2}}}}{{{\textstyle{{mx} \over 2}}}}} \right\}}^2}\,\,\,\frac{{{m^2}{x^2}}}{4}.\frac{1}{{{{\left\{ {\frac{{\sin {\textstyle{{nx} \over 2}}}}{{{\textstyle{{nx} \over 2}}}}} \right\}}^2}}}.\frac{4}{{{n^2}{x^2}}}} \right]$

$ = \frac{{{m^2}}}{{{n^2}}} \times 1 = \frac{{{m^2}}}{{{n^2}}}$.

Aliter : Apply $L$-Hospital’s rule,

$\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos mx}}{{1 - \cos nx}} = \mathop {\lim }\limits_{x \to 0} \frac{{m\sin mx}}{{n\sin nx}} $

$= \mathop {\lim }\limits_{x \to 0} \frac{{{m^2}\cos mx}}{{{n^2}\cos nx}} = \frac{{{m^2}}}{{{n^2}}}.$

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

Aliter : Apply $L-$ Hospital’s rule,

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

$ = \mathop {\lim }\limits_{x \to \infty } \frac{{\sqrt x \,(5)}}{{\sqrt x \,\left( {\sqrt {1 + \frac{5}{x}} + 1} \right)}} = \frac{5}{2}$.

Apply $L-$ Hospital‘s rule, 

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

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