Question

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

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

Answer 1

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

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

Replace $n$ with $\frac{1}{x}$,

Now as $n\to \infty ,x\to 0$

$\therefore$ Limit becomes

$L=\underset{x\to 0}{lim}{\left(\right(x+1{)}^{-\alpha }+\sin x)}^{\frac{1}{x}}$

Taking $\log$ on both sides, we get

$\ln L=\underset{x\to 0}{lim}\frac{\ln \left(\right(x+1{)}^{-\alpha }+\sin x)}{x}$

It os forming the $\frac{0}{0}$ form,

So, using L'Hospital's Rule, we get

$\ln L=\underset{x\to 0}{lim}\frac{-\alpha (x+1{)}^{-\alpha -1}+\cos x}{(x+1{)}^{-\alpha }+\sin x}$

$\Rightarrow \ln L=1-\alpha$

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