Question

Question no. 1 akshita165

Answer 1

$\displaystyle Q1.\ \ \lim_{n\to\infty}\ \dfrac{n^{^p}\sin^2(n!)}{n+1},\ \ p\in(0,\ 1)$

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$n\to\infty\ \ \&\ \ p\in(0,\ 1)\ \ \implies\ \ n^{^p}\in(1,\ n)\\ \\ \therefore\ n^{^p}\to n\ \ \implies\ \ n^{^p}\to\infty$

$\sin^2(n!)\in[0,\ 1]\ \ \Longleftarrow\ \ \sin(n!)\in[-1,\ 1]\ \ \text{even}\ \ n!\longrightarrow\infty\\ \\ \therefore\ \sin^2(n!)\to 1$

$\&\ \ n\to\infty\ \ \implies\ \ n+1\to\infty$

$\displaystyle\therefore\ \lim_{n\to\infty}\dfrac{n^{^p}\sin^2(n!)}{n+1}\ =\ \dfrac{\infty}{\infty}$

$\boxed{\begin{minipage}{11.44cm}\text{L'Hospital's Rule:}\\ \\ \begin{center} \displaystyle\lim_{n\to\infty}\frac{f}{g}\ =\ \lim_{n\to\infty}\frac{f'}{g'}\ \iff\ \lim_{n\to\infty}\frac{f}{g}\ =\ \frac{0}{0} \ OR \ \dfrac{\pm\infty}{\pm\infty} \end{center}\end{minipage}}$

$f=n^{^p}\sin^2(n!)\ \ \Longrightarrow\ \ f'=pn^{^{p-1}}\sin^2(n!)+n^{^p}\ \!2\sin(n!)\cos(n!)\\ \\ g=n+1\ \ \implies\ \ g'=1$

$\displaystyle\lim_{n\to\infty}\dfrac{n^{^p}\sin^2(n!)}{n+1}\ =\ \lim_{n\to\infty}\dfrac{pn^{^{p-1}}\sin^2(n!)+n^{^p}\ \!2\sin(n!)\cos(n!)}{1}\\ \\ \\ =\ \lim_{n\to\infty}pn^{^{p-1}}\sin^2(n!)+n^{^p}\ \!2\sin(n!)\cos(n!)$

$\displaystyle p\in(0,\ 1)\ \ \implies\ \ p-1\in(-1,\ 0).\ \ \therefore\ \ pn^{^{p-1}}\to\ 1\\ \\ \\ \sin(n!)\ \&\ \cos(n!)\in[-1,\ 1].\ \ \therefore\ \ 2\sin(n!)\cos(n!)\to 2\\ \\ \\ \therefore\ \lim_{n\to\infty}pn^{^{p-1}}\sin^2(n!)+n^{^p}\ \!2\sin(n!)\cos(n!)\ =\ 1+2\cdot\infty\ \to\ \infty$

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$\displaystyle Q2.\ \ \lim_{n\to\infty}(0.2)^{\log_{\sqrt3}(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{2^{n+1}})}$

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$\displaystyle\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+...+\dfrac{1}{2^{n+1}}\ =\ \sum_{k=1}^{n}\frac{1}{2^{k+1}}\ \ \text{is a Geometric Series.}\\ \\ \\ \boxed{\displaystyle\sum_{k=1}^{n}ar^{k-1}\ =\ \dfrac{a(r^n-1)}{r-1}}\\ \\ \\ \text{Here,}\ \ a=\dfrac{1}{4}\ \ \ ;\ \ \ r\ =\ \dfrac{1}{8}\div\dfrac{1}{4}\ =\ \dfrac{1}{2}$

$\displaystyle\therefore\ \sum_{k=1}^n\frac{1}{2^{k+1}}\ =\ \dfrac{\frac{1}{4}((\frac{1}{2})^n-1)}{\frac{1}{2}-1}\ =\ \dfrac{\frac{1}{4}(1-\frac{1}{2^n})}{1-\frac{1}{2}}\\ \\ \\ =\ \dfrac{\frac{1}{4}(1-\frac{1}{2^n})}{\frac{1}{2}}\ =\ \dfrac{1}{2}\left(1-\dfrac{1}{2^n}\right)\ =\ \dfrac{1}{2}-\dfrac{1}{2^{n+1}}$

$\displaystyle\therefore\ \lim_{n\to\infty}(0.2)^{\log_{\sqrt3}(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{2^{n+1}})}\\ \\ \\ =\ 0.2^{^{\displaystyle\small\text{ \lim_{n\to\infty}\log_{\sqrt{3}}\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{2^{n+1}}\right) }}}\\ \\ \\ =\ 0.2^{^{\displaystyle\small\text{ \lim_{n\to\infty}\log_{\sqrt3}\left(\dfrac{1}{2}-\dfrac{1}{2^{n+1}}\right) }}}$

$n\to\infty\ \ \implies\ \ 2^{^{n+1}}\to\infty\ \ \implies\ \ \dfrac{1}{2^{^{n+1}}}\ \to\ 0$

$\therefore\ 0.2^{^{\displaystyle\small\text{ \lim_{n\to\infty}\log_{\sqrt3}\left(\dfrac{1}{2}-\dfrac{1}{2^{n+1}}\right) }}}\\ \\ \\ =\ 0.2^{^{\displaystyle\small\log_{\sqrt3}\left(\dfrac{1}{2}-0\right)}}\\ \\ \\ =\ 0.2^{^{\displaystyle\small\text{ \log_{\sqrt{3}}\dfrac{1}{2} }}}$

$=\ 0.2^{^{\displaystyle\small\text{ \frac{\log\frac{1}{2}}{\log\sqrt3} }}}\ =\ 0.2^{^{\displaystyle\small\text{ \frac{-\log 2}{\frac{1}{2}\log3} }}}\\ \\ \\ =\ \left(0.2^{^{\displaystyle\small-2}}\right)^{\displaystyle\small\dfrac{\log2}{\log3}}\ =\ 25^{^{\displaystyle\small\log_32}}\ \approx\ 8$

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$\displaystyle\smallQ3.\ \ \lim_{n\to\infty}\dfrac{1}{1\cdot4}+\dfrac{1}{4\cdot7}+\dfrac{1}{7\cdot10}+...+\dfrac{1}{(3n-2)(3n+1)}\ =\ \lim_{n\to\infty}\sum_{k=1}^{n}\frac{1}{(3k-2)(3k+1)}$

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$\boxed{\displaystyle\sum_{k=1}^{n}\dfrac{1}{(3k-2)(3k+1)}\ =\ \dfrac{n}{3n+1}}$

$\displaystyle\therefore\ \lim_{n\to\infty}\sum_{k=1}^n\frac{1}{(3k-1)(3k+2)}\ =\ \lim_{n\to\infty}\frac{n}{3n+1}\ =\dfrac{\infty}{\infty}\\ \\ \\ \therefore\ \lim_{n\to\infty}\frac{n}{3n+1}\ =\ \lim_{n\to\infty}\frac{1}{3}\ =\ \dfrac{1}{3}$