Question

$\underset{n\to \infty }{lim}{\{{\left(\frac{n}{n+1}\right)}^{\alpha }+\sin \frac{1}{n}\}}^n(\text{where}\alpha \in N)$ is equal to

• A. $e^{-\alpha }$
• B. $-\alpha$
• C. $e^{1-\alpha }$
• D. $e^{1+\alpha }$

Answer 1

The correct option is C $e^{1-\alpha }$

Given limit is of $1^{\infty }$ form
$\Rightarrow L=e^{\underset{n\to \infty }{lim}n({\left(\frac{n}{n+1}\right)}^{\alpha }+\sin \frac{1}{n}-1)}$

$=e^{\underset{n\to \infty }{lim}n\sin \frac{1}{n}+\underset{n\to \infty }{lim}n({\left(\frac{n}{n+1}\right)}^{\alpha }-1)}$
Let $L_1=\underset{n\to \infty }{lim}n\sin \frac{1}{n}$
$=\underset{n\to \infty }{lim}\frac{\sin \frac{1}{n}}{1/n}=1$

Let $L_2=\underset{n\to \infty }{lim}n({\left(\frac{n}{n+1}\right)}^{\alpha }-1)$
$=\underset{n\to \infty }{lim}n\left(\frac{n^{\alpha }-(1+n{)}^{\alpha }}{(n+1{)}^{\alpha }}\right)$

$=\underset{n\to \infty }{lim}n\left(\frac{n^{\alpha }-(n^{\alpha }+{}^{\alpha }{C}_1{n}^{\alpha -1}+{}^{\alpha }{C}_2{n}^{\alpha -2}+{}^{\alpha }{C}_3{n}^{\alpha -3}+\dots )}{(n+1{)}^{\alpha }}\right)$

$=\underset{n\to \infty }{lim}n\left(\frac{-n^{\alpha -1}(\alpha +{}^{\alpha }{C}_2\frac{1}{n}+{}^{\alpha }{C}_3\frac{1}{n^2}+\dots )}{n^{\alpha }{(1+\frac{1}{n})}^{\alpha }}\right)$
$L_2=-\alpha$

$\therefore L=e^{1-\alpha }$