Maths M.C.Q (1 Marks)

$\lim _{x \rightarrow \infty} x^3 \times\left\{\frac{x^3\left\{\left(\sqrt{3+\frac{1}{x}}+\sqrt{3-\frac{1}{x}}\right)^6+\left(\sqrt{3+\frac{1}{x}}-\sqrt{3-\frac{1}{x}}\right)^6\right\}}{x^6\left\{\left(1+\sqrt{1-\frac{1}{x^2}}\right)^6+\left(1-\sqrt{1-\frac{1}{x^2}}\right)^6\right\}}\right\}$

$=\frac{(2 \sqrt{3})^6+0}{2^6+0}=3^3=(27)$

$\lim _{x \rightarrow 0} 2\left(\frac{1}{2} \times \frac{9}{1} \times \frac{1 \times 64}{1 \times 32}\right)=18$

$\therefore x^2+b x+a=a(1-\alpha x)(1-\beta x)$

$\therefore \lim _{x \rightarrow \frac{1}{\alpha}}\left\{\frac{1-\cos \left(x^2+b x+a\right)}{2(1-\alpha x)^2}\right\}^{\frac{1}{2}}=\lim _{x \rightarrow \frac{1}{2}}\left\{\frac{1-\cos a(1-\alpha x)(1-\beta x)}{2\{a(1-\alpha x)(1-\beta x)\}^2} \cdot a^2(1-\beta x)^2\right\}^{\frac{1}{2}}$

$=\left[\frac{1}{2} \cdot \frac{1}{2} a ^2\left(1-\frac{\beta}{\alpha}\right)^2\right]^{\frac{1}{2}}$

$=\frac{1}{2} \frac{1}{\alpha \beta}\left(1-\frac{\beta}{\alpha}\right)=\frac{1}{2}\left(\frac{1}{\alpha \beta}-\frac{1}{\alpha^2}\right)$

$=\frac{1}{2 \alpha}\left(\frac{1}{\beta}-\frac{1}{\alpha}\right)=\frac{1}{ k }\left(\frac{1}{\beta}-\frac{1}{\alpha}\right)$

$\therefore k =2 \alpha$

On expansion,

$\lim _{x \rightarrow 0} \frac{\left(1+a x+\frac{a^2 x^2}{2}\right)-\left(1-\frac{b^2 x^2}{2}\right)-\frac{c x}{2}(1-c x)}{2 x^2}=17$

$\lim _{x \rightarrow 0} \frac{\left(a-\frac{c}{2}\right) x+x^2\left(\frac{a^2}{2}+\frac{b^2}{2}+\frac{c^2}{2}\right)}{2 x^2}=17$

For limit to be exist a $-\frac{c}{2}=0$

$a=\frac{c}{2}$

and $\frac{a^2+b^2+c^2}{4}=17$

$a^2+b^2+4 a^2=17 \times 4$

$5 a^2+b^2=68$

On rationalising each term

$\lim _{n \rightarrow \infty} \sqrt{\frac{d}{n}}\left(\frac{\sqrt{a_n}-\sqrt{a_1}}{d}\right)$

$\lim _{n \rightarrow \infty} \sqrt{\frac{d}{n}}\left(\frac{(n-1) d}{\left(\sqrt{a_n}+\sqrt{a_1}\right) d}\right)=1$

$--\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n+1}}\right) < \left(2^{\frac{1}{2}}-2^{\frac{1}{2 n+1}}\right)^n$

$\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)^n < L < \left(2^{\frac{1}{2}}-2^{\frac{1}{2 n+1}}\right)^n$

$\lim _{n \rightarrow \infty}\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)^n=0 \text { and } \lim _{n \rightarrow \infty}\left(2^{\frac{1}{2}}-2^{\frac{1}{2 n+1}}\right)^n=0$

$\Rightarrow \lim _{n \rightarrow \infty} L=0$

$\lim \limits_{x \rightarrow 0^{-}} f(x)=\alpha-\frac{4}{3} \Rightarrow 0+\frac{\alpha^{-1}-2}{-1}=\alpha-\frac{4}{3}$

$\Rightarrow 2-\frac{1}{\alpha}=\alpha-\frac{4}{3}$

$\Rightarrow \alpha+\frac{1}{\alpha}=\frac{10}{3}$

$\Rightarrow \alpha=3 ; \alpha \in I$

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

$\therefore k =1$

$=\lim \limits_{x \rightarrow \frac{\pi}{2}} \frac{\tan ^{2} x\left[\sin ^{2} x-3 \sin x+2\right]}{\sqrt{9}+\sqrt{9}}$

$=\lim \limits_{x \rightarrow \frac{\pi}{2}} \frac{\tan ^{2} x(\sin x-1)(\sin x-2)}{6}$

$=\frac{1}{6} \lim \limits_{x \rightarrow \frac{\pi}{2}} \tan ^{2} x(1-\sin x)$

$=\frac{1}{6} \lim \limits_{x \rightarrow \frac{\pi}{2}} \frac{\sin ^{2} x(1-\sin x)}{(1-\sin x)(1+\sin x)}=\frac{1}{12}$

$\lim \limits_{x \rightarrow 0}\left(\frac{2 \cdot \sin \left(\frac{x+\sin x}{2}\right) \sin \left(\frac{x-\sin x}{2}\right)}{x^{4}}\right)$

$\lim\limits _{x \rightarrow 0} 2\left(\frac{\sin \left(\frac{x+\sin x}{2}\right)}{\left(\frac{x+\sin x}{2}\right)}\left(\frac{\sin \left(\frac{x-\sin x}{2}\right)}{\left(\frac{x-\sin x}{2}\right)}\right)\left(\frac{\left.\frac{x+\sin x}{2}\right)}{x^{4}}\left(\frac{x-\sin x}{2}\right)\right.\right.$

$\lim\limits _{x \rightarrow 0}\left(\frac{x^{2}-\sin ^{2} x}{2 x^{4}}\right):\left(\frac{0}{0}\right)$

Apply $L-Hopital\; Rule$ :

$\lim \limits_{x \rightarrow 0} \frac{2 x-2 \sin x \cos x}{2.4 \cdot x^{3}}$

$\lim\limits _{x \rightarrow 0} \frac{2 x-\sin 2 x}{8 x^{3}} ; \frac{0}{0}:$ Again apply $L-Hopital \;rule$

$\lim \limits_{x \rightarrow 0} \frac{2-2 \cos (2 x)}{8(3) x^{2}}$

$\lim \limits_{x \rightarrow 0} \frac{2(1-\cos (2 x))}{24\left(4 x^{2}\right)} \times 4 \Rightarrow \frac{2}{24} \times \frac{1}{2} \times 4 \Rightarrow \frac{1}{6}$

$\lim \limits_{x \rightarrow \frac{1}{\sqrt{2}}} \frac{\sin \left(\sin ^{-1} \sqrt{1-x^{2}}\right)-x}{1-\tan \left(\tan ^{-1}\left(\frac{\sqrt{1-x^{2}}}{x}\right)\right)}$

$\lim \limits_{x \rightarrow \frac{1}{\sqrt{2}}} \frac{\sqrt{1-x^{2}}-x}{1-\left(\frac{\sqrt{1-x^{2}}}{x}\right)}$

$\lim \limits_{x \rightarrow \frac{1}{\sqrt{2}}}(-x)=-\frac{1}{\sqrt{2}}$

$L.H.L.$ $\lim \limits_{x \rightarrow 7-} \frac{18-[1-x]}{[x]-3 a}$

$=\frac{18-(-6)}{6-3 a}$

$=\frac{24}{6-3 a}$

$R.H.L.$ $\lim \limits_{x \rightarrow 7^{+}} \frac{18-[1-x]}{[x]-3 a}$

$=\frac{18-(-7)}{7-3 a}$

$=\frac{25}{7-3 a}$

Now $L.H.L. \;= \;R.H.L.$

$\frac{24}{6-3 a}=\frac{25}{7-3 a}$

$\Rightarrow 168-72 a=150-75 a$

$\Rightarrow 18=-3 a$

$\Rightarrow a =-6$

For finite limit

$a+b-5=0$...............$(1)$

Apply $L'H $rule

$\lim\limits _{x \rightarrow 1} \frac{\cos \left(3 x^{2}-4 x+1\right)(6 x-4)-2 x}{\left(6 x^{2}-14 x+a\right)}=-2$

For finite limit

$6-14+a=0$

$a=8$

From $(1)\,\,\,\,\,\, b=-3$

Now $(a-b)=11$

$=\lim _{x \rightarrow \frac{\pi}{4}} \frac{-7(\cos x+\sin x)^{6}(-\sin x+\cos x)}{-2 \sqrt{2} \cos 2 x}$ using $L-H$

$=\lim _{x \rightarrow \frac{\pi}{4}} \frac{56(\cos x-\sin x)}{2 \sqrt{2} \cos 2 x}\left(\frac{0}{0}\right)$

$=\lim _{x \rightarrow \frac{\pi}{4}} \frac{-56(\sin x+\cos x)}{-4 \sqrt{2} \sin 2 x} \quad$ using L-H Rule

$=7 \sqrt{2} \cdot \sqrt{2}=14$

$\lim _{n \rightarrow \infty}\left\{1-\frac{1}{2}\left(\frac{n+1}{n^{2}}\right)+\frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2 !}\left(\frac{n+1}{n^{2}}\right)^{2}+\ldots .\right\}+\alpha n+\beta=0$

$\lim _{n \rightarrow \infty} n-\frac{1}{2}+\frac{1}{n}+\ldots+n \alpha+\beta=0$

$\alpha=-1, \beta=\frac{1}{2}$

$8(\alpha+\beta)=-4$

$\lim _{n \rightarrow \infty} \frac{(n+1)^{k-1}}{n^{k+1}}[n k \cdot n+1+2+\ldots+n]$

$=\lim _{n \rightarrow \infty} \frac{(n+1)^{k-1}}{n^{k+1}} \cdot\left[n^{2} k+\frac{n(n+1)}{2}\right]$

$=\lim _{n \rightarrow \infty} \frac{(n+1)^{k-1} \cdot n^{2}\left(k+\frac{\left(1+\frac{1}{n}\right)}{2}\right)}{n^{k+1}}$

$\lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)\left(k+\frac{\left(1+\frac{1}{n}\right)}{2}\right)$

$\left(k+\frac{1}{2}\right)$

$RHS$

$\lim _{ n \rightarrow \infty} \frac{1}{ n ^{ k +1}}\left(1^{ k }+2^{ k }+\ldots+ n ^{ k }\right)=\frac{1}{ k +1}$

$\text { LHS }=\text { RHS }$

$k +\frac{1}{2}=33 \cdot \frac{1}{ k +1}$

$(2 k +1)( k +1)=66$

$( k -5)(2 k +13)=0$

$k =5 \text { or }-\frac{13}{2}$

$\beta=\lim _{x \rightarrow 0} \frac{1+\alpha x-\left[1+3 x+\frac{9 x^{2}}{2 !}+\ldots .\right]}{(\alpha x) \frac{\left(e^{3 x}-1\right)}{3 x} 3 x}$

$\beta=\lim _{x \rightarrow 0} \frac{(\alpha x-3 x)-\frac{9 x^{2}}{2 !}-\ldots \ldots .}{3 \alpha x^{2}}$

For existence of limit $\alpha-3=0$ $\alpha=3$

Limit $\beta=\frac{-3}{2 \alpha}$

$\beta=-\frac{1}{2}$

Now,$\alpha+\beta=\frac{5}{2}$

From $1^{\infty}$

$=e^{\lim _{x \rightarrow 0}\left[\left(\frac{(x+2 \cos x)^{3}+2(x+2 \cos x)^{2}+3 \sin (x+2 \cos x)}{(x+2)^{3}+2(x+2)^{2}+3 \sin (x+2)}\right)-1\right] \times \frac{100}{x}}$

$=e^{\lim _{x \rightarrow 0}\left[\frac{100}{x}\left(\frac{(x+2 \cos x)^{3}+2(x+2 \cos x)^{2}+3 \sin (x+2 \cos x)-\left((x+2)^{3}+2(x+2)^{2}+3 \sin (x+2)\right)}{(x+2)^{3}+2(x+2)^{2}+3 \sin (x+2)}\right)\right]}$

$=e^{\lim _{x \rightarrow 0} \frac{100}{x}\left[\left(\frac{(x+2 \cos x)^{3}+(x+2)^{3}+2(x+2 \cos x)^{2}-2(x+2)^{2}+3 \sin (x+2 \cos x)-3 \sin (x+2)}{8+8+3 \sin ^{2}}\right)\right]}$

$=e^{\frac{100}{16+3 \sin ^{2}}} \lim _{x \rightarrow 0} \frac{3(x+2 \cos x)^{2} \times(1+2 \sin x)-3(x+2)^{2}-4(x+2 \cos x)}{x(1-2 \sin x)-4(x+2)+3 \cos (x+2 \cos x) \times(1-2 \sin x)-3 \cos (x+2)}$

$=e^{\frac{100}{16+3 \sin 2}}\left(\frac{12-3(4)+8 \times 1-8+3 \cos 2-3 \cos 2}{1}\right)$

Using $L'H$ rule.

$=e^{o}=1$

constant terms should be zero

$a+\beta=0$

coeff of $x$ should be zero

$\alpha-\beta+\gamma=0$

$\operatorname{coeff}$ of $x^{2}$ should be zero

$\lim _{x \rightarrow 0} \frac{x^{3}\left(\frac{\alpha}{3 !}-\frac{\beta}{3 !}-\frac{\gamma}{3 !}\right)+x^{4}\left(\frac{\alpha}{3 !}-\frac{\beta}{3 !}-\frac{\gamma}{3 !}\right)}{x}=\frac{2}{3}$

$\frac{\alpha}{2}+\frac{\beta}{2}=0$

$\frac{\alpha}{6}-\frac{\beta}{6}-\frac{\gamma}{6}=2 / 3$

$\alpha=1, \beta=-1, \gamma=-2$

$=\lim _{\theta \rightarrow 0} \frac{-\tan \left(\pi \sin ^{2} \theta\right)}{\sin \left(2 \pi \sin ^{2} \theta\right)}$

$=\lim _{\theta \rightarrow 0}-\left(\frac{\tan \left(\pi \sin ^{2} \theta\right)}{\pi \sin ^{2} \theta}\right)\left(\frac{2 \pi \sin ^{2} \theta}{\sin \left(2 \pi \sin ^{2} \theta\right)}\right) \times \frac{1}{2}$

$=\frac{-1}{2}$ 

$=\lim _{n \rightarrow a} \tan \left(\sum_{r=1}^{n} \tan ^{-1}\left(\frac{r+1-r}{1+r(r+1)}\right)\right)$

$=\tan \left(\lim _{n \rightarrow a} \sum_{r=1}^{n}\left[\tan ^{-1}(r+1)-\tan ^{-1}(r)\right]\right)$

$=\tan \left(\lim _{n \rightarrow \infty}\left(\tan ^{-1}(n+1)-\frac{\pi}{4}\right)\right)$

$=\tan \left(\frac{\pi}{4}\right)=1$

$\lim _{x \rightarrow a} \frac{x f(a)-a f(x)}{x-a}$

$\Rightarrow \lim _{x \rightarrow a} \frac{f(a)-a f^{\prime}(x)}{1} \quad$ (Lopitals rule)

$=f(a)-a f^{\prime}(a)$

$=4-2 a$

$S=\sum_{n=1}^{9} \frac{2}{4\left(n^{2}+3 n+2\right)}=\frac{1}{2} \sum_{n=1}^{9}\left(\frac{1}{n+1}-\frac{1}{n+2}\right)$

$S=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{11}\right)=\frac{9}{44}$

$\lim _{x \rightarrow 2}\left(\frac{2 x f(2)-4 f^{\prime}(x)}{1}\right)$

$=\frac{4(4)-4}{1}=12$

$L=\lim _{h \rightarrow 0} \frac{4 \sinh }{3 h}$

$\Rightarrow L=\frac{4}{3}$

$\Rightarrow \lim _{x \rightarrow 0} \frac{a\left(1+x+\frac{x^{2}}{2 !} \ldots\right)-b\left(1-\frac{x^{2}}{2 !}+\ldots\right)+c\left(1-x+\frac{x^{2}}{2 !}\right)}{\left(\frac{x \sin x}{x}\right) x}=2$

$a-b+c=0$  $.......(1)$

$a-c=0$  $............(2)$

$\frac{a+b+c}{2}=2$

$\Rightarrow a+b+c=4$

So $6 L+1=2$

and

$\begin{array}{c} r \leq[ r ]< r +1 \\ 2 r \leq[2 r ]<2 r +1 \end{array} $$ $$ \begin{array}{ccc} 3 r & \leq[3 r ] & <3 r +1 \\ \vdots & \vdots & \vdots \\ nr & \leq[ nr ] & < nr +1 \end{array}$

$r +2 r +\ldots+ nr \leq[ r ]+[2 r ]+\ldots+[ nr ]<( r +2 r +\ldots+ nr )+ n$

$\frac{\frac{n(n+1)}{2} \cdot r}{n^{2}} \leq \frac{[r]+[2 r]+\ldots . .+[n r]}{n^{2}}<\frac{\frac{n(n+1)}{2} r+n}{n^{2}}$

Now,

$\lim _{n \rightarrow \infty} \frac{n(n+1) \cdot r}{2 \cdot n^{2}}=\frac{r}{2}$

and

$\lim _{n \rightarrow \infty} \frac{\frac{n(n+1) r}{2}+n}{n^{2}}=\frac{r}{2}$

So, by Sandwich Theorem, we can conclude that

$\lim _{n \rightarrow \infty} \frac{[r]+[2 r]+\ldots \ldots+[n r]}{n^{2}}=\frac{r}{2}$

$\text { So, } l=\exp \left(\lim _{n \rightarrow \infty} \frac{1+\frac{1}{2}+\frac{1}{3}+\ldots \ldots .+\frac{1}{n}}{n}\right)$

Now,

$0 \leq 1+\frac{1}{2}+\frac{1}{3}+\ldots .+\frac{1}{n} \leq 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}$

$\leq 2 \sqrt{n}-1$

So, $l=\exp (0)$ (from sandwich theorem)

$=1$

$=\lim _{x \rightarrow 0} \frac{\operatorname{ax}-\left(e^{4 x}-1\right)}{\operatorname{ax} \cdot 4 x} \quad$ Use $\lim _{x \rightarrow 0} \frac{e^{4 x}-1}{4 x}=1$

Apply L'Hospital Rule

$=\lim _{x \rightarrow 0} \frac{a-4 e ^{4 x }}{8 ax } \quad\left(\frac{ a -4}{0}\right.$ form $)$

limit exists only when $a-4=0 \Rightarrow a=4$

$=\lim _{x \rightarrow 0} \frac{4-4 e^{4 x}}{32 x}$

$=\lim _{x \rightarrow 0} \frac{1-e^{4 x}}{8 x}$

$\left(\frac{0}{0}\right)$

$=\lim _{x \rightarrow 0} \frac{- e ^{4 x } \cdot 4}{8}=-\frac{1}{2} \Rightarrow b =-\frac{1}{2}$

$a-2 b=4-2\left(-\frac{1}{2}\right)$

$=5$

$\Rightarrow \lim _{x \rightarrow \beta} \frac{1\left(1+\frac{2\left(x^{2}+b x+c\right)}{1 !}+\frac{2^{2}\left(x^{2}+b x+c\right)^{2}}{2 !}+\ldots\right)-1-2\left(x^{2}+b x+c\right)}{(x-\beta)^{2}}$

$\Rightarrow \lim _{x \rightarrow \beta} \frac{2\left(x^{2}+b x+1\right)^{2}}{(x-\beta)^{2}}$

$\Rightarrow \lim _{x \rightarrow \beta} \frac{2(x-\alpha)^{2}(x-\beta)^{2}}{(x-\beta)^{2}}$

$\Rightarrow 2(\beta-\alpha)^{2}=2\left(b^{2}-4 c\right)$

$\Rightarrow a>0$

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

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

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

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

$\Rightarrow \frac{-1}{1+a}=b \Rightarrow b=-\frac{1}{2}$

$(a, b)=\left(1,-\frac{1}{2}\right)$

$\lim _{x \rightarrow 0} \frac{1-\cos \left(2 \pi \cos ^{4} x\right)}{2 x^{4}}$

$\lim _{x \rightarrow 0} \frac{1-\cos \left(2 \pi-2 \pi \cos ^{4} x\right)}{\left[2 \pi\left(1-\cos ^{4} x\right)\right]^{2}} 4 \pi^{2} \cdot \frac{\sin ^{4} x}{2 x^{4}}\left(1+\cos ^{2} x\right)^{2}$

$=\frac{1}{2} \cdot 4 \pi^{2} \cdot \frac{1}{2}(2)^{2}=4 \pi^{2}$

Using L Hopital rule

$\alpha=\lim _{x \rightarrow \frac{\pi}{4}} \frac{3 \tan ^{2} x \sec ^{2} x-\sec ^{2} x}{-\sin \left(x+\frac{\pi}{4}\right)}$

$\Rightarrow \alpha=-4$

$\beta=\lim _{x \rightarrow 0}(\cos x)^{\cot x}=e^{\lim _{x \rightarrow 0} \frac{(\cos x-1)}{\tan x}}$

$\beta=e^{\lim _{x \rightarrow 0} \frac{-(1-\cos x)}{x^{2}} \frac{x^{2}}{\left(\frac{\tan x}{x}\right)^{x}}}$

$\beta=e^{\lim _{x \rightarrow 0}\left(\frac{-1}{2}\right)^{\frac{x}{1}}}=e^{0} \Rightarrow \beta=1$

$\alpha=-4 ; \beta=1$

If $a x^{2}+b x-4=0$ are the roots then $16 a-4 b-4=0\, \&\, a+b-4=0$

$\Rightarrow \mathrm{a}=1 \,\&\, \mathrm{~b}=3$

$\lim _{x \rightarrow 1} \frac{x^{n} f(1)-f(x)}{x-1}=44$

$\lim _{x \rightarrow 1} \frac{9 x^{n}-\left(x^{6}+2 x^{4}+x^{3}+2 x+3\right)}{x-1}=44$

$\lim _{x \rightarrow 1} \frac{9 n x^{n-1}-\left(6 x^{5}+8 x^{3}+3 x^{2}+2\right)}{1}=44$

$\Rightarrow 9 n-(19)=44$

$\Rightarrow 9 n=63$

$\Rightarrow n=7$

form $1^{\infty}$

$e^{\lim _{x \rightarrow 0}}\left(\frac{1-\cos x \sqrt{\cos 2} x}{x^{2}}\right) \times(x+2)$

Now limt ${x \rightarrow 0} \frac{\lim _{x \rightarrow 0} \frac{1-\cos x \sqrt{\cos 2} x}{}}{x^{2}}$

$\lim _{x \rightarrow 0} \frac{\sin x \sqrt{\cos 2} x-\cos x \times \frac{1}{2 \sqrt{\cos 2 x}} \times(-2 \sin 2 x)}{x^{2}}$

(by L' Hospital Rule)

$\lim _{x \rightarrow 0} \frac{\sin x \cos 2 x+\sin 2 x \cdot \cos x}{2 x}$

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

So, $e^{\operatorname{limit_x \rightarrow 0}\left(\frac{1-\cos x \sqrt{\cos 2 x}}{x^{2}}\right)(x+2)}$

$=e^{\frac{3}{2} \times 2}=e^{3}$

$\Rightarrow a=3$

$\lim _{x \rightarrow 0} \frac{x(\alpha-\beta)+x^{2}\left(\alpha+\frac{\beta}{2}+\gamma\right)+x^{3}\left(\frac{\alpha}{2}-\frac{\beta}{3}-\gamma\right)}{x^{3}}$

For limit to exist

$\alpha-\beta=0, \alpha+\frac{\beta}{2}+\gamma=0$

$\frac{\alpha}{2}-\frac{\beta}{3}-\gamma=10 \ldots \text { (i) }$

$\beta=\alpha, \gamma=-3 \frac{\alpha}{2}$

Put in $(i)$

$\frac{\alpha}{2}-\frac{\alpha}{3}+\frac{3 \alpha}{2}=10$

$\frac{\alpha}{6}+\frac{3 \alpha}{2}=10 \Rightarrow \frac{\alpha+9 \alpha}{6}=10$

$\Rightarrow \alpha=6$

$\alpha=6, \beta=6, \gamma=-9$

$\alpha+\beta+\gamma=3$

$\lim _{x \rightarrow 0} \frac{x\left\{(1-\sin x)^{1 / 8}+(1+\sin x)^{1 / 8}\right\}\left\{(1-\sin x)^{1 / 4}+(1+\sin x)^{1 / 4}\right\}\left\{(1-\sin x)^{1 / 2}+(1+\sin x)^{1 / 2}\right\}}{(1-\sin x-1-\sin x)}$

$\lim _{x \rightarrow 0} \frac{8 x}{-2 \sin x}=-4$

Divide by $3^{2 r }$

$\sum_{r=1}^{k} \tan ^{-1}\left(\frac{\left(\frac{2}{3}\right)^{r}}{\left(\frac{2}{3}\right)^{2 r} \cdot 2+3}\right)$

$\sum_{r=1}^{k} \tan ^{-1}\left(\frac{\left(\frac{2}{3}\right)^{r}}{3\left(\left(\frac{2}{3}\right)^{2 r+1}+1\right)}\right)$

Let $\left(\frac{2}{3}\right)^{r}=t$

$\sum_{r=1}^{k} \tan ^{-1}\left(\frac{\frac{t}{3}}{1+\frac{2}{3} t^{2}}\right)$

$\sum_{r=1}^{k} \tan ^{-1}\left(\frac{t-\frac{2 t}{3}}{1+t \cdot \frac{2 t}{3}}\right)$

$\sum_{ r =1}^{ k }\left(\tan ^{-1}( t )-\tan ^{-1}\left(\frac{2 t }{3}\right)\right)$

$\sum_{ r =1}^{ k }\left(\tan ^{-1}\left(\frac{2}{3}\right)^{ r }-\tan ^{-1}\left(\frac{2}{3}\right)^{ r +1}\right)$

$S _{ k }=\tan ^{-1}\left(\frac{2}{3}\right)-\tan ^{-1}\left(\frac{2}{3}\right)^{ k +1}$

$S _{\infty}=\lim _{ k \rightarrow \infty}\left(\tan ^{-1}\left(\frac{2}{3}\right)-\tan ^{-1}\left(\frac{2}{3}\right)^{ k +1}\right)$

$=\tan ^{-1}\left(\frac{2}{3}\right)-\tan ^{-1}(0)$

$\therefore \quad S _{\infty}=\tan ^{-1}\left(\frac{2}{3}\right)=\cot ^{-1}\left(\frac{3}{2}\right)$

$\Rightarrow \quad \lim _{x \rightarrow 1}\left(\frac{x-1}{x-1}+\frac{x^{2}-1}{x-1}+\ldots . . \frac{x^{n}-1}{x-1}\right)=820$

$\Rightarrow 1+2+\ldots .+n=820$

$\Rightarrow \quad n(n+1)=2 \times 820$

$\Rightarrow \quad n(n+1)=40 \times 41$

since $\mathrm{n} \in \mathrm{N},$ so $n=40$

$=\lim \limits_{x \rightarrow 0} \frac{1}{x}\left\{\tan \left(\frac{\pi}{4}+x\right)-1\right\}$

$=\lim \limits_{x \rightarrow 0}\left(\frac{1+\tan x-1+\tan x}{x(1-\tan x)}\right)$

$=e^{\lim \limits_{x \rightarrow 0} \frac{2 \tan x}{x(1-\tan x)}}$

$=e^{2}$

$\Rightarrow \lim _{x \rightarrow 0} \frac{\left(1-\cos \frac{x^{2}}{2}\right)\left(1-\cos \frac{x^{2}}{4}\right)}{4\left(\frac{x^{2}}{2}\right)^{2}}=\frac{1}{16\left(\frac{x^{2}}{4}\right)^{2}}=\frac{1}{8} \times \frac{1}{32}=2^{-k}$

$\Rightarrow 2^{-8}=2^{-k} \Rightarrow k=8$

$\mathrm{k}=3+2=5$

$=\lim _{x \rightarrow 2} \frac{\left(3^{x}-9\right)\left(3^{x}-3\right)}{\left(3^{x / 2}-3\right)}$

$=\lim _{x \rightarrow 2} \frac{\left(3^{x / 2}+3\right)\left(3^{x / 2}-3\right)\left(3^{x}-3\right)}{\left(3^{x / 2}-3\right)}$

$=36$

$=e^{\lim _{x \rightarrow 0}\left(\frac{-4}{7 x^{2}+2}\right)}=\frac{1}{e^{2}}$

$f(\mathrm{x})=\left[\mathrm{x}^{2}\right] \sin (\pi \mathrm{x})$ will be discontinuous at nonintegers

$\therefore \mathrm{x}=\sqrt{\mathrm{A}+1}$ i.e. $\sqrt{5}$

$ L =\lim _{h \rightarrow 0} \frac{(a+2(a+h))^{1 / 3}-(3(a+h))^{1 / 3}}{(3 a+a+h)^{1 / 3}-(4(a+h))^{1 / 3}} $

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

$=\lim _{h \rightarrow 0}\left(\frac{3^{1 / 3}}{4^{1 / 3}}\right)\left[\frac{\left(1+\frac{2 h}{9 a}\right)-\left(1+\frac{h}{3 a}\right)}{\left(1+\frac{h}{12 a}\right)-\left(1+\frac{h}{3 a}\right)}\right]$

$=\left(\frac{3}{4}\right)^{1 / 3} \frac{\left(\frac{2}{9}-\frac{1}{3}\right)}{\left(\frac{1}{12}-\frac{1}{3}\right)}=\left(\frac{3}{4}\right)^{1 / 3}\left(\frac{8-12}{3-12}\right)$

$=\left(\frac{3}{4}\right)^{1 / 3}\left(\frac{-4}{-9}\right)=\frac{4^{1-\frac{1}{3}}}{3^{2-\frac{1}{3}}}=\frac{4^{2 / 3}}{3^{5 / 3}}$

$=\frac{(8 \times 2)^{1 / 3}}{(27 \times 9)^{1 / 3}}=\frac{2}{3}\left(\frac{2}{9}\right)^{1 / 3}$

$\operatorname{RHL}: \lim _{x \rightarrow 0^{+}}\left|\frac{1-x+x}{\lambda-x+1}\right|=\left|\frac{1}{\lambda}\right|$

For existence of limitt

$LHL = RHD$

$\Rightarrow \frac{1}{|\lambda-1|}=\frac{1}{|\lambda|} \Rightarrow \lambda=\frac{1}{2}$

$\therefore \quad L=\frac{1}{|\lambda|}=2$