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Question Number 8066 by Chantria last updated on 29/Sep/16

$$calculate$$$$\underset{n\to \mathrm{\infty }}{lim}\frac{n}{2^{\sqrt{n}}}$$$$$$

Answered by prakash jain last updated on 02/Oct/16

$$2^{\sqrt{n}}=e^{\sqrt{n}\ln 2}=1+\sqrt{n}\ln 2+\frac{{\left(\sqrt{n}\ln 2\right)}^2}{2!}+\frac{{\left(\sqrt{n}\ln 2\right)}^3}{3!}+..$$$$\frac{n}{2^{\sqrt{n}}}=\frac{n}{1+\sqrt{n}\ln 2+\frac{{\left(\sqrt{n}\ln 2\right)}^2}{2!}+\frac{{\left(\sqrt{n}\ln 2\right)}^3}{3!}+..}$$$$=\frac{\frac{n}{n}}{1+\frac{\ln \sqrt{2}}{n}+\frac{\ln 2}{2!}+\frac{\sqrt{\mathrm{n}}\ln 2}{3!}+\left(\mathrm{higher}\mathrm{power}\mathrm{of}n\right)}$$$$=\frac{1}{1+\frac{\ln \sqrt{2}}{n}+\frac{\ln 2}{2!}+\frac{\sqrt{\mathrm{n}}\ln 2}{3!}+\left(\mathrm{higher}\mathrm{power}\mathrm{of}n\right)}$$$$\underset{n\to \mathrm{\infty }}{\lim }\frac{n}{2^{\sqrt{n}}}=\frac{1}{1+0+\frac{\ln 2}{2!}+\mathrm{\infty }}=0$$

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