Thank you for reporting, we will resolve it shortly
Q.
$\displaystyle\lim _{n \rightarrow \infty}\left(\left(\frac{n}{n+1}\right)^{\alpha}+\sin \frac{1}{n}\right)^{n} \quad($ where $\alpha \in Q)$ is equal to
Limits and Derivatives
Solution:
\displaystyle
$1^{\infty}$ form
$L = e ^{\displaystyle\lim _{n \rightarrow \infty} n\left(\left(\frac{n}{n+1}\right)^{\alpha}+\sin \frac{1}{n}-1\right)}$
$= e ^{\displaystyle\lim _{n \rightarrow \infty} n \sin \frac{1}{n}+\displaystyle\lim _{n \rightarrow \infty} n\left(\left(\frac{n}{n+1}\right)^{\alpha}-1\right)}$
Consider, $\displaystyle\lim _{n \rightarrow \infty} n\left(\left(\frac{n}{n+1}\right)^{\alpha}-1\right)$
$=\displaystyle\lim _{n \rightarrow \infty} n\left(\left(\frac{1}{1+1 / n}\right)^{\alpha}-1\right)$
Put $n =\frac{1}{ y }$
$=\displaystyle\lim _{y \rightarrow 0} \frac{1}{y}\left(\left(\frac{1}{1+y}\right)^{\alpha}-1\right)$
$=\displaystyle\lim _{y \rightarrow 0} \frac{1-(1+y)^{\alpha}}{y}=-\alpha$
(Using binomial)
$\therefore L = e ^{1-\alpha}$