Question Number 63921 by raj last updated on 11/Jul/19
$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\left(\frac{{n}}{{n}+\mathrm{1}}\right)^{\alpha} +\mathrm{sin}\:\frac{\mathrm{1}}{{n}}\right)^{{n}} \mathrm{where}\:\alpha\in\mathbb{Q} \\ $$$$\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$
Commented by kaivan.ahmadi last updated on 11/Jul/19
$$\sim{lim}_{{n}\rightarrow\infty} \left(\left(\frac{{n}}{{n}+\mathrm{1}}\right)^{\alpha} −\mathrm{1}+{sin}\frac{\mathrm{1}}{{n}}\right)\left({n}\right)= \\ $$$${lim}_{{n}\rightarrow\infty} \left(\frac{{n}^{\alpha+\mathrm{1}} −{n}\left({n}+\mathrm{1}\right)^{\alpha} }{\left({n}+\mathrm{1}\right)^{\alpha} }+{nsin}\frac{\mathrm{1}}{{n}}\right)= \\ $$$$\sim{lim}_{{n}\rightarrow\infty} \left(\frac{−\alpha{n}^{\alpha} }{{n}^{\alpha\:} }+{n}×\frac{\mathrm{1}}{{n}}\right)=−\alpha+\mathrm{1} \\ $$
Commented by raj last updated on 11/Jul/19
$${thank}\:{you} \\ $$