Question

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

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

Answer 1

Answer: (C)

$e^{1-\alpha }$

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

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

$\text{Put}n=\frac{1}{y}\text{. Then}$

$\underset{y\to 0}{lim}\frac{1}{y}({\left(\frac{1}{1+y}\right)}^{\alpha }-1)=\underset{y\to 0}{lim}\frac{1-(1+y{)}^{\alpha }}{y}=-a\text{(Using binomial)}\therefore L=e^{1-\alpha }$