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Question Number 63921 by raj last updated on 11/Jul/19

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

Commented by kaivan.ahmadi last updated on 11/Jul/19

$$\sim {lim}_{n\to \mathrm{\infty }}({\left(\frac{n}{n+1}\right)}^{\alpha }-1+sin\frac{1}{n})\left(n\right)=$$$${lim}_{n\to \mathrm{\infty }}(\frac{n^{\alpha +1}-n{(n+1)}^{\alpha }}{{(n+1)}^{\alpha }}+nsin\frac{1}{n})=$$$$\sim {lim}_{n\to \mathrm{\infty }}(\frac{-\alpha n^{\alpha }}{n^{\alpha }}+n\times \frac{1}{n})=-\alpha +1$$

Commented by raj last updated on 11/Jul/19

$$thankyou$$

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