Maths M.C.Q (1 Marks)

$ = {e^{\mathop {\lim }\limits_{x \to \infty } \left[ {\frac{{ - 4\left( {3 - \frac{1}{x}} \right)}}{{\left( {1 - \frac{1}{x}} \right)}}} \right]}} = {e^{ - 12}}$.

Using $ L-$ Hospital’s rule three times, then

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

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

$ = 1$.

$ = \mathop {\lim }\limits_{n \to \infty } \frac{{{{(10)}^n}\left[ {{{\left( {\frac{1}{{10}}} \right)}^n} - 1} \right]}}{{{{(10)}^{n + 1}}\left( {1 + \frac{1}{{{{10}^{n + 1}}}}} \right)}} = - \frac{1}{{10}}$

$\therefore \alpha = 1$.

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

[By using $ L-$  Hospital rule]

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

$ = \frac{1}{{1 + 0}}.{(1)^2}.\frac{1}{1} = 1$.

$\left. { + \left( {\frac{1}{{(2n - 1)}} - \frac{1}{{(2n + 1)}}} \right)} \right]$

$ = \mathop {\lim }\limits_{n \to \infty } \frac{1}{2}\left[ {1 - \frac{1}{{2n + 1}}} \right] = \frac{1}{2}$.

$ = \mathop {\lim }\limits_{n \to \infty } \,\,\frac{1}{6}\frac{{n\,(n + 1)\,(2n + 1)}}{{1 + {n^3}}}$

$ = \mathop {\lim }\limits_{n \to \infty } \,\frac{1}{6}\frac{{\left( {1 + \frac{1}{n}} \right)\,\left( {2 + \frac{1}{n}} \right)}}{{\left( {\frac{1}{{{n^3}}} + 1} \right)}}$

$ = \frac{1}{6}\,.\,1\,.\,\frac{2}{{(1)}} = \left( {\frac{1}{3}} \right).$

$ = \mathop {\lim }\limits_{n \to \infty } \,\frac{{2\,(1 - n)}}{n}$

$ = \mathop {\lim }\limits_{n \to \infty } 2\,\left( {\frac{1}{n} - 1} \right)$$ = 2(0 - 1) = - 2$.

$ = \mathop {\lim }\limits_{x \to \infty } \frac{{{x^{10}}\left[ {{{\left( {1 + \frac{1}{x}} \right)}^{10}} + {{\left( {1 + \frac{2}{x}} \right)}^{10}} + ... + {{\left( {1 + \frac{{100}}{x}} \right)}^{10}}} \right]}}{{{x^{10}}\left[ {1 + \frac{{{{10}^{10}}}}{{{x^{10}}}}} \right]}} = 100$.

$ = \mathop {\lim }\limits_{n \to \infty } \frac{{n(n + 1)}}{{2({n^2} + 100)}} = \mathop {\lim }\limits_{n \to \infty } \frac{{{n^2}\left( {1 + \frac{1}{n}} \right)}}{{2{n^2}\left( {1 + \frac{{100}}{{{n^2}}}} \right)}} = \frac{1}{2}$.

Applying  $ L- $ Hospital rule, we get

$\mathop {\lim }\limits_{x \to 0} \frac{{\int_0^x {\cos {t^2}dt} }}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{\cos {x^2}}}{1} = 1$.

$ = \frac{1}{3}\,\left[ {\frac{{2\sqrt 2 }}{2}} \right] = \frac{{\sqrt 2 }}{3},\,\,0 < \frac{{\sqrt 2 }}{3} < \frac{1}{2}.$

Aliter : Apply $ L-$ Hospital’s rule,

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

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

$ = \frac{{\frac{1}{2} + \frac{1}{2}}}{{\frac{3}{{2\sqrt 2 }} + \frac{3}{{2\sqrt 2 }}}} = \frac{{2\sqrt 2 }}{6} = \frac{{\sqrt 2 }}{3}$.

$ = \mathop {\lim }\limits_{h \to 0} \,\,(1 - (1 - h) + [1 - h - 1] + [1 - (1 - h)])\,$

$ = \mathop {\lim }\limits_{h \to 0} \,(h + [ - h] + [h]) = \mathop {\lim }\limits_{h \to 0} \,(h - 1 + 0) = - 1$

and $\mathop {\lim }\limits_{x \to 1 + } \,(1 - x + [x - 1] + [1 - x])\,$

$ = \mathop {\lim }\limits_{h \to 0} \,(1 - (1 + h) + [1 + h - 1] + [1 - (1 + h)])$

$ = \mathop {\lim }\limits_{h \to 0} \,( - h + [h] + [ - h]) = \mathop {\lim }\limits_{h \to 0} \,( - h + 0 - 1) = - 1$

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

$ - 1 = \mathop {\lim }\limits_{x \to a} \,\,\frac{{{a^x} - {x^a}}}{{{x^x} - {a^a}}} = \mathop {\lim }\limits_{x \to a} \,\frac{{{a^x}{{\log }_e}a - a{x^{a - 1}}}}{{{x^x} + {a^a}{{\log }_e}a}}$

$ \Rightarrow \,\, - 1 = \frac{{{a^a}{{\log }_e}a - a\,.\,{a^{a - 1}}}}{{{a^a} + {a^a}{{\log }_e}a}} = \frac{{{{\log }_e}a - 1}}{{{{\log }_e}a + 1}}$.....$(i)$

Now $(i)$ is satisfied only when $a = 1.$

$I$. $\lim _{n \rightarrow \infty} \frac{2^n+(-2)^n}{2^n}=\lim _{n \rightarrow \infty} 1+(-1)^n=$ does not exist

$11$. $\lim _{n \rightarrow \infty} \frac{3^n+(-3)^n}{4^n}=\lim _{n \rightarrow \infty}\left(\frac{3}{4}\right)^n+\left(\frac{-3}{4}\right)^n$

$=0+0=0$

Given, $S_n=\sum_{k=\bullet}^n \frac{1}{\sqrt{n^2+k}}$

$\frac{1}{\sqrt{n^2+n}}+\frac{1}{\sqrt{n^2+n}}+\ldots+\frac{1}{\sqrt{n^2+n}}$

$\leq \frac{1}{\sqrt{n^2}}+\frac{1}{\sqrt{n^2+1}}+\ldots+\frac{1}{\sqrt{n^2+n}}$ $\leq \frac{1}{\sqrt{n^2}}+\frac{1}{\sqrt{n^2}}+\ldots+\frac{1}{\sqrt{n^2}}$

$\lim _{n \rightarrow \infty} \frac{n+1}{\sqrt{n^2+n}} \leq \lim _{n \rightarrow \infty} S_n \leq \lim _{n \rightarrow \infty} \frac{n+1}{\sqrt{n^2}}$

$1 \leq \lim _{n \rightarrow \infty} S_n \leq 1 \Rightarrow \lim _{n \rightarrow \infty} S_n=1$

We have, $\lim _{x \rightarrow-\infty}\left(\sqrt{4 x^2-x}+2 x\right)$

$=\lim _{x \rightarrow-\infty} \frac{4 x^2-x-4 x^2}{\sqrt{4 x^2-x-2 x}}$

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

$=\lim _{x \rightarrow-\infty} \frac{1}{\sqrt{4-\frac{1}{x}+2}}$

We have,

$x_n =\left(2^n+3^n\right)^{\frac{1}{2 n}}$

$\Rightarrow \quad \lim _{n \rightarrow \infty} x_n =\lim _{n \rightarrow \infty}\left(2^n+3^n\right)^{\frac{1}{2 n}}$

$=\lim _{n \rightarrow \infty} 3^{n / 2 n}\left[\left(\frac{2}{3}\right)^n+1\right]^{\frac{1}{2 n}}$

$=3^{3 / 2} \times 1=\sqrt{3}$

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

Apply L-Hospital rule

$L=\lim _{x \rightarrow 2} \frac{a^x \log a-a^{3-x} \log a}{-a^{3-x} \log a-\frac{1}{2} e^{x / 2} \log a}$

$L=\lim _{x \rightarrow 2} \frac{a^x-a^{3-x}}{-a^{3-x}-\frac{a^{x / 2}}{2}}$

$L=\frac{a^2-a}{-a-\frac{a}{2}}=\frac{2}{3}(1-a)$

Let

$p =\lim _{x \rightarrow 0}\left(\frac{x}{\sin x}\right)^{6 / x^2}$

$\Rightarrow \quad \log p =\lim _{x \rightarrow 0} \frac{6}{x^2} \log \left(\frac{x}{\sin x}\right)$

$\Rightarrow \quad \log p =\lim _{x \rightarrow 0} \frac{6 \log \left(\frac{x}{\sin x}\right)}{x^2}$

Apply $L-$ Hospital rule

$\log p=\lim _{x \rightarrow 0}$ 

$6 \frac{\frac{\sin x}{x} \frac{(\sin x-x \cos x)}{\sin ^2 x}}{2 x}$

$\log p=\lim _{x \rightarrow 0} 6 \frac{\sin x(\sin x-x \cos x)}{x \cdot 2 x \sin ^2 x}$

$\log p=\lim _{x \rightarrow 0} 3 \frac{\sin x}{x} \times \lim _{x \rightarrow 0}$

$\frac{\sin x-x \cos x}{\frac{\sin ^2 x}{x^2}} \times x^3$

$\log p=3 \times \lim _{x \rightarrow 0} \frac{\sin x-x \cos x}{x^3}$

$\Rightarrow \quad \log p=3 \times \lim _{x \rightarrow 0} \frac{\sin x-x \cos x}{x^3}$

Again apply L.Hospital's rule

$\log p=3 \lim _{x \rightarrow 0} \frac{\cos x-\cos x+x \sin x}{3 x^2}$

$\Rightarrow \log p=\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$

$\therefore \quad p=e^1=e$

Let

$p =\lim _{x \rightarrow 0}\left(\frac{x}{\sin x}\right)^{6 / x^2}$

$\log p =\lim _{x \rightarrow 0} \frac{6}{x^2} \log \left(\frac{x}{\sin x}\right)$

$\log p =\lim _{x \rightarrow 0} \frac{6 \log \left(\frac{x}{\sin x}\right)}{x^2}$ 

Apply L-Hospital rule

$\log p=\lim _{x \rightarrow 0} 6 \frac{\frac{\sin x}{x} \frac{(\sin x-x \cos x)}{\sin ^2 x}}{2 x}$

$\Rightarrow \log p=\lim _{x \rightarrow 0} 6 \frac{\sin x(\sin x-x \cos x)}{x \cdot 2 x \sin ^2 x}$

$\Rightarrow \log p=\lim _{x \rightarrow 0} 3 \frac{\sin x}{x} \times \lim _{x \rightarrow 0}$

$\frac{\sin x-x \cos x}{\frac{\sin ^2 x}{x^2}} \times x^3$

$\Rightarrow \quad \log p=3 \times \lim _{x \rightarrow 0} \frac{\sin x-x \cos x}{x^3}$

Again apply L.Hospital's rule

$\quad \log p=3 \lim _{x \rightarrow 0} \frac{\cos x-\cos x+x \sin x}{3 x^2}$

$\Rightarrow \log p=\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$

$\therefore \quad p=e^1=e$

$\lim _{x \rightarrow \infty} f(x)=M > 0$

(a) $\lim _{x \rightarrow \infty} x \sin \left(\frac{1}{x}\right) f(x)=\lim _{x \rightarrow \infty} \frac{\sin \left(\frac{1}{x}\right)}{\frac{1}{x}} f(x)$ $=\lim _{x \rightarrow \infty} \frac{\sin \left(\frac{1}{x}\right)}{\frac{1}{x}} \times \lim _{x \rightarrow \infty} f(x)=1 \times M=M$

Hence, option (a) is true.

(b) $\lim _{x \rightarrow \infty} \sin (f(x))$

$\sin \left(\lim _{x \rightarrow \infty} f(x)\right)=\sin M$

Option (b) is also true.

(c) $\lim _{x \rightarrow \infty} x \sin \left(e^{-x}\right) f(x)$

$\lim _{x \rightarrow \infty} x e^{-x} \frac{\sin \left(e^{-x}\right)}{e^{-x}} f(x)$

$=0 \times 1 \times M=0$

Hence, option (c) is false.

(d) $\lim _{x \rightarrow \infty} \frac{\sin x}{x} f(x)=0 \times M=0$

Option (d) is also true.

We have,

$a_n = a _1 \cdot a_2 \cdot a_3 \ldots a_{n-1}+4, a_1=5$

$a_2 =a_1+4=5+4=9$

$a_3 =a_1 a_2+4=5 \times 9+4=49$

$a_4 =a_1 a_2 a_3=5 \times 9 \times 49+4=2209$

$a_4=\left(a_3-2\right)^2=(49-2)^2=47^2=2209$

$a_3 =\left(a_2-2\right)^2=(9-2)^2=49$

$a_n =\left(a_{n-1}-2\right)^2$

$\therefore \quad \lim _{n \rightarrow \infty} \frac{\sqrt{ a _n}}{a_{n-1}} =\lim _{n \rightarrow \infty} \frac{\sqrt{\left(a_{n-1}-2\right)^2}}{a_{n-1}}$

$=\lim _{n \rightarrow \infty}\left(\frac{ a _{n-1}-2}{a_{n-1}}\right)=1$

We have,

$a_n=\sqrt{\frac{1+a_{n-1}}{2}}$ for $n \geq 1, a_0=\cos \theta$

$a_1=\sqrt{\frac{1+a_0}{2}}=\sqrt{\frac{1+\cos \theta}{2}}$

$a_1=\sqrt{\frac{2 \cos ^2 \frac{\theta}{2}}{2}}=\cos \frac{\theta}{2}$

$a_2=\sqrt{\frac{1+\cos \frac{\theta}{2}}{2}}=\cos \frac{\theta}{2^2}$

$a_n=\cos \frac{\theta}{2^n}$

$\lim _{n \rightarrow \infty} 4^n\left(1-a_n\right)$

$=\lim _{n \rightarrow \infty} 4^n\left(1-\cos \frac{\theta}{2^n}\right)$

$=\lim _{n \rightarrow \infty} 4^n\left(2 \sin ^2 \frac{\theta}{2^{n+1}}\right)$

$=\lim _{n \rightarrow \infty} 4^n \cdot 2\left(\frac{\sin \frac{\theta}{2^{n+1}}}{\frac{\theta}{2^{n+1}}}\right)^2 \times \frac{\theta^2}{2^{2 n+2}}$

$=\lim _{n \rightarrow \infty} \frac{4^n}{4^n \cdot 2}\left(\frac{\sin \frac{\theta}{2^{n+1}}}{\frac{\theta}{2^{n+1}}}\right)^2 \times \theta^2=\frac{\theta^2}{2}$

We have all natural numbers whose decimal expansion has only even digits $0,2,4,6,8$.

$a_n=n$th number of sequence if these are arranged in increasing orders.

Let $n=5^K$, then $5$ based representation of $n$ is a $1$ followed by $K$ zeroes,

$\therefore$ Decimal representation and doubling its value, we obtain a $2$ followed by $K$ zeroes that is $210^K$.

$\therefore \quad \log a_n=K+\log _{10} 2$

$\Rightarrow \log n =K \log _{10} 5$

$\Rightarrow \quad \lim _{n \rightarrow \infty} \frac{\log a_n}{\log n} =\frac{K+\log _{10} 2}{K \log _{10} 5}$

$K$ goes to infinity

$\therefore \quad \lim _{n \rightarrow \infty} \frac{\log a_n}{\log n}=\frac{1}{\log _{10} 5}=\log _5 10$

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

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

Applying limit $\mathrm{a}=\frac{1}{4 \sqrt{2}}$

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

$ b=\lim _{x \rightarrow 0}(1+\cos x)(\sqrt{2}+\sqrt{1+\cos x})$

$ \Rightarrow \lim _{x \rightarrow 0} \frac{3+\alpha\left[x-\frac{x^3}{3 !}+\ldots\right]+\beta\left[1-\frac{x^2}{2 !}+\frac{x^4}{4 !} \ldots\right]+\left(-x-\frac{x^2}{2}-\frac{x^3}{3} \ldots\right)}{3 \tan ^2 x}=\frac{1}{3} $

$ \Rightarrow \lim _{x \rightarrow 0} \frac{(3+\beta)+(\alpha-1) x+\left(-\frac{1}{2}-\frac{\beta}{2}\right) x^2+\ldots}{3 x^2} \times \frac{x^2}{\tan ^2 x}=\frac{1}{3} $

$ \Rightarrow \beta+3=0, \alpha-1=0 \text { and } \frac{-\frac{1}{2}-\frac{\beta}{2}}{3}=\frac{1}{3} $

$ \Rightarrow \beta=-3, \alpha=1 $

$ \Rightarrow 2 \alpha-\beta=2+3=5$

$\therefore a=1$

$\mathrm{b}=\lim _{\mathrm{x} \rightarrow 0} \frac{\int_0^{\mathrm{x}} \frac{\ln (1+\mathrm{t})}{1+\mathrm{t}^{2024}} \mathrm{dt}}{\mathrm{x}^2}$

Using $L' HOPITAL$ Rule

$b=\lim _{x \rightarrow 0} \frac{\ln (1+x)}{\left(1+x^{2024}\right)} \times \frac{1}{2 x}=\frac{1}{2}$

$\begin{array}{ll}\text { Now, } \mathrm{cx}^2+\mathrm{dx}+\mathrm{e}=0, & \mathrm{x}^2+\mathrm{x}+4=0 \\ & (\mathrm{D}<0)\end{array}$

$\therefore \frac{\mathrm{c}}{1}=\frac{\mathrm{d}}{1}=\frac{\mathrm{e}}{4}$

$lim _{x \rightarrow 0} \frac{e^{2 s i n x}-2|\sin x|-1}{|\sin x|^2} \times \frac{\sin ^2 x}{x^2}$

Let $|\sin x|=t$

$\lim _{t \rightarrow 0} \frac{e^{2 t}-2 t-1}{t^2} \times \lim _{x \rightarrow 0} \frac{\sin ^2 x}{x^2}$

$=\lim _{t \rightarrow 0} \frac{2 e^{2 t}-2}{2 t} \times 1=2 \times 1=2$

$=\lim _{x \rightarrow \infty} \frac{(c-b) x+\left(\frac{b}{2}-c+a\right) x^2+\left(a-\frac{b}{3}+\frac{c}{2}\right) x^3+\ldots \ldots .}{x^3}=1$

$\mathrm{c}-\mathrm{b}=0, \quad \frac{\mathrm{b}}{2}-\mathrm{c}+\mathrm{a}=0$

$ \mathrm{a}-\frac{\mathrm{b}}{3}+\frac{\mathrm{c}}{2}=1 \quad \mathrm{a}=\frac{3}{4} \quad \mathrm{~b}=\mathrm{c}=\frac{3}{2} $

$ \mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2=\frac{9}{16}+\frac{9}{4}+\frac{9}{4} $

$16\left(\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2\right)=81$

$=\lim _{x \rightarrow \infty} \frac{(c-b) x+\left(\frac{b}{2}-c+a\right) x^2+\left(a-\frac{b}{3}+\frac{c}{2}\right) x^3+\ldots \ldots .}{x^3}=1$

$\mathrm{c}-\mathrm{b}=0, \quad \frac{\mathrm{b}}{2}-\mathrm{c}+\mathrm{a}=0$

$ \mathrm{a}-\frac{\mathrm{b}}{3}+\frac{\mathrm{c}}{2}=1 \quad \mathrm{a}=\frac{3}{4} \quad \mathrm{~b}=\mathrm{c}=\frac{3}{2} $

$ \mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2=\frac{9}{16}+\frac{9}{4}+\frac{9}{4} $

$16\left(\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2\right)=81$

$16\left(\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2\right)=81$

$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)$

$=\lim _{h \rightarrow 0} f(h)$

$=\lim _{h \rightarrow 0} \frac{\cos ^{-1}\left(1-h^2\right) \sin ^{-1}(1-h)}{h\left(1-h^2\right)}$

$=\lim _{h \rightarrow 0} \frac{\cos ^{-1}\left(1-h^2\right)}{h}\left(\frac{\sin ^{-1} 1}{1}\right)$

Let $\cos ^{-1}\left(1-h^2\right)=\theta \Rightarrow \cos \theta=1-h^2$

$=\frac{\pi}{2} \lim _{\theta \rightarrow 0} \frac{\theta}{\sqrt{1-\cos \theta}}$

$=\frac{\pi}{2} \lim _{\theta \rightarrow 0} \frac{1}{\sqrt{\frac{1-\cos \theta}{\theta^2}}}$

$=\frac{\pi}{2} \frac{1}{\sqrt{1 / 2}}$

$\mathrm{R}=\frac{\pi}{\sqrt{2}}$

$=\frac{8}{3} \frac{\sqrt{5}}{6^{2 / 3}} \quad \begin{gathered}\mathrm{m}=8 \\ \mathrm{n}=3\end{gathered}$

$ 8 m+12 n=100$

Using L Hopital Rule.

$\lim _{x \rightarrow 0} \frac{f^{\prime}(x)}{3 x^2}=\lim _{x \rightarrow 0} \frac{x+\sin \left(1-e^x\right)}{3 x^2}$ (Again L Hopital)

Using $L.H.$ Rule

$ =\lim _{x \rightarrow 0} \frac{-\left[\sin \left(1-e^x\right)\left(-e^x\right) \cdot e^x+\cos \left(1-e^x\right) \cdot e^x\right]}{6} $

$ =-\frac{1}{6}$

$ =16 \times \frac{2}{\mathrm{a}^2}\left(\frac{1}{\mathrm{a}}-\frac{1}{\mathrm{~b}}\right)^2 $

$ =\frac{32}{\mathrm{a}^2}\left(\frac{17}{4}\right)=\frac{17.8}{\mathrm{a}^2}=\frac{17 \times 8 \times 16}{(-1+\sqrt{117})^2} $

$ =\frac{136.16}{18.2 \sqrt{7}} \times \frac{18+2 \sqrt{7}}{18+2 \sqrt{7}} $

$ =\frac{136}{256}(18+2 \sqrt{7}) \cdot 16 $

$ =153+17 \sqrt{17}=\alpha+\beta \sqrt{17} $

$ \alpha+\beta=153+17=170$

$ \lim _{n \rightarrow \infty} \frac{\sum_{r=1}^{n-1}\left(-r^3+r^2(n+1)-n r\right)}{\left(\frac{n(n+1)}{2}\right)^2-\frac{n(n+1)(2 n+1)}{6}} $

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

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

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

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

By expansion

$\lim _{x \rightarrow 0} \frac{2\left(1-\left(1-\frac{x^2}{2}\right)\right)\left(1-\frac{1}{2} \cdot \frac{4 x^2}{2}\right)\left(1-\frac{1}{3} \cdot \frac{9 x^2}{2}\right) \ldots \ldots\left(1-\frac{1}{10} \cdot \frac{100 x^2}{2}\right)}{x^2}$

$\lim _{x \rightarrow 0} 2\left(\frac{1-\left(1-\frac{x^2}{2}\right)\left(1-\frac{2 x^2}{2}\right)\left(1-\frac{3 x^2}{2}\right) \ldots\left(1-\frac{10 x^2}{2}\right)}{x^2}\right)$

$\lim _{x \rightarrow 0} \frac{2\left(1-1+x^2\left(\frac{1}{2}+\frac{2}{2}+\frac{3}{2}+\ldots .+\frac{10}{2}\right)\right)}{x^2}$

$ 2\left(\frac{1}{2}+\frac{2}{2}+\frac{3}{2}+\ldots . .+\frac{10}{2}\right) $

$ 1+2+\ldots \ldots+10=\frac{10 \times 11}{2}=55$

$ =1 $

$ \beta=\lim _{x \rightarrow 0}(1+\sin x)^{\frac{1}{2} \cot x} $

$ =e^{1 / 2} $

$ x^2-(1+\sqrt{e})+\sqrt{e}=0 $

$ a x^2+b x-\sqrt{e}=0$

On comparing

$ a=-1, b=\sqrt{e}+1 $

$ 12 \ln (a+b)=12 \times \frac{1}{2}=6$

$ \mathrm{x}=\frac{1-\mathrm{c}^2}{2+\mathrm{c}-3 \mathrm{c}^2}, \mathrm{y}=\frac{1-3 \mathrm{x}}{5}=\frac{\mathrm{c}-1}{5\left(2+\mathrm{c}-3 \mathrm{c}^2\right)} $

$ \mathrm{h}=\lim _{\mathrm{c} \rightarrow 1} \frac{(1-\mathrm{c})(1+\mathrm{c})}{(1-\mathrm{c})(2+3 \mathrm{c})}=\frac{2}{5} $

$ \mathrm{~K}=\lim _{\mathrm{c} \rightarrow 1} \frac{\mathrm{c}-1}{-5(\mathrm{c}-1)(3 \mathrm{c}+2)}=-\frac{1}{25} $

$ \text { Centre }\left(\frac{2}{25},-\frac{1}{25}\right), $

$ \mathrm{r}=\sqrt{\left(2-\frac{2}{5}\right)^2+\left(0-\frac{1}{25}\right)^2}=\sqrt{\frac{64}{25}+\frac{1}{625}} $

$ \mathrm{r}=\frac{\sqrt{161}}{25} $

$ \left(\mathrm{x}-\frac{2}{5}\right)^2+\left(\mathrm{y}+\frac{1}{25}\right)^2=\frac{161}{125} $

$ \Rightarrow 25 \mathrm{x}^2+25 \mathrm{y}^2-20 \mathrm{x}+2 \mathrm{y}-60=0$

$ =\operatorname{Lim}_{x \rightarrow 0}(-e) \frac{\left(e^{\frac{\ln (1+2 x)}{2 x}-1}-1\right)}{x} $

$ =\operatorname{Lim}_{x \rightarrow 0}(-e) \frac{\ln (1+2 x)-2 x}{2 x^2} $

$ =(-e) \times(-1) \frac{4}{2 \times 2}=e$

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

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

$ =\left(1(1)+\frac{1}{2}\right) 3\left(\frac{\pi}{2}\right)^2 $

$ =\frac{9 \pi^2}{8}$

$\lim _{x \rightarrow a}([x]-5-[2 x]-2)=0$

$\lim _{x \rightarrow a}([x]-[2 x])=7$

${[a]-[2 a]=7}$

$a \in I, \quad a=-7$

$a \notin I, \quad a=I+f$

$\text { Now, }[a]-[2 a]=7$

$\quad-I-[2 f]=7$

$\text { Case-I: f } \in\left(0, \frac{1}{2}\right)$

$2 f \in(0,1)$

$-I=7$

$I=-7 \Rightarrow a \in(-7,-6.5)$

Case$-II:$ $f \in\left(\frac{1}{2}, 1\right)$

$2 f \in(1,2)$

$- I -1=7$

$I =-8 \Rightarrow a \in(-7.5,-7)$

Hence, $a \in(-7.5,-6.5)$

$\operatorname{Lim}_{n \rightarrow \infty} \frac{3 n(n-1)}{2\left(\sqrt{2 n^4+4 n+3}-\sqrt{n^4+5 n+4}\right)}$

$=\frac{3}{2(\sqrt{2}-1)}=\frac{3}{2}(\sqrt{2}+1)$

$\lim _{h \rightarrow 0} \frac{1}{2}\left(\frac{(2 p+h)^2-4 p(2 p+h)+q^2+82+16}{h^2}\right)^2=\frac{1}{2}$

Using L'Hospital's

$\lim _{x \rightarrow 2 p^{+}}[f(x)]=0$

Applying L'Hospitals Rule

$48 \lim _{x \rightarrow 0} \frac{x^3}{x^6+1} \times \frac{1}{4 x^3}$

$=12$