Question Number 585 by 123456 last updated on 02/Feb/15
$${f}_{{n}} \left({x}\right)=\left(\mathrm{1}+\frac{{x}}{{n}}\right)^{{nx}} ,{n}\in\mathbb{N}^{\ast} ,{x}\in\mathbb{R},{x}\geqslant\mathrm{0} \\ $$$$\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\:{f}_{{n}} \left({x}\right)= \\ $$
Answered by prakash jain last updated on 02/Feb/15
$${y}=\left(\mathrm{1}+\frac{{x}}{{n}}\right)^{{nx}} \\ $$$$\mathrm{ln}\:{y}={nx}\mathrm{ln}\:\left(\mathrm{1}+\frac{{x}}{{n}}\right)=\frac{{x}\mathrm{ln}\:\left(\mathrm{1}+\frac{{x}}{{n}}\right)}{\frac{\mathrm{1}}{{n}}} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}ln}\:{y}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}{x}\centerdot\frac{\frac{−\frac{{x}}{{n}^{\mathrm{2}} }}{\mathrm{1}+\frac{{x}}{{n}}}}{−\frac{\mathrm{1}}{{n}^{\mathrm{2}} }}=\frac{{x}^{\mathrm{2}} }{\mathrm{1}+\frac{{x}}{{n}}}={x}^{\mathrm{2}} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{f}_{{n}} \left({x}\right)={e}^{{x}^{\mathrm{2}} } \\ $$