Limit Continuity And Differentiability Ques 44
- If $f:[-1,1] \rightarrow R$ and $f^{\prime}(0)=\lim _{n \rightarrow \infty} n f\left(\frac{1}{n}\right)$ and $f(0)=0$.
Find the value of $\lim _{n \rightarrow \infty} \frac{2}{\pi}(n+1) \cos ^{-1}\left(\frac{1}{n}\right)-n$, given that
$ 0<\left|\lim _{n \rightarrow \infty} \cos ^{-1}\left(\frac{1}{n}\right)\right|<\frac{\pi}{2} \text {. } $
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Answer:
Correct Answer: 44.$\left(1-\frac{2}{\pi}\right)$
Solution: Here, $ \lim _{n \rightarrow \infty} \frac{2}{\pi}(n+1) \cos ^{-1}\left(\frac{1}{n}\right)-n $
$ = \lim _{n \rightarrow \infty} n\left\{\frac{2}{\pi}\left(1+\frac{1}{n}\right) \cos ^{-1}\left(\frac{1}{n}\right)-1\right\}=\lim _{n \rightarrow \infty} n f\left(\frac{1}{n}\right) $
where, $f\left(\frac{1}{n}\right)=\frac{2}{\pi}\left(1+\frac{1}{n}\right) \cos ^{-1}\left(\frac{1}{n}\right)-1=f^{\prime}(0)$
$ \left[\text { given, } f^x(0)=\lim _{n \rightarrow \infty } n f\left(\frac{1}{n}\right)\right] $
$\therefore \quad \lim _{n \rightarrow \infty } \frac{2}{\pi}(n+1) \cos ^{-1} \frac{1}{n}-n=f^{\prime}(0)$ $\quad$ ……..(i)
where, $f(x)=\frac{2}{\pi}(1+x) \cos ^{-1} x-1, f(0)=0$
$\Rightarrow \quad f^{\prime}(x)=\frac{2}{\pi}\left\{(1+x) \frac{-1}{\sqrt{1-x^2}}+\cos ^{-1} x\right\}$
$\Rightarrow \quad f^{\prime}(0)=\frac{2}{\pi}\left\{-1+\frac{\pi}{2}\right\}=1-\frac{2}{\pi}$ $\quad$ ……..(ii)
$\therefore \quad$ From Eqs. (i) and (ii), we get
$ \lim _{n \rightarrow \infty } \frac{2}{\pi}(n+1) \cos ^{-1}\left(\frac{1}{n}\right)-n=1-\frac{2}{\pi} $
BETA
