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Question Number 585 by 123456 last updated on 02/Feb/15

f_n (x)=(1+(x/n))^(nx) ,n∈N^∗ ,x∈R,x≥0  lim_(n→+∞)  f_n (x)=

$${f}_{{n}} \left({x}\right)=\left(\mathrm{1}+\frac{{x}}{{n}}\right)^{{nx}} ,{n}\in\mathbb{N}^{\ast} ,{x}\in\mathbb{R},{x}\geqslant\mathrm{0} \\ $$$$\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\:{f}_{{n}} \left({x}\right)= \\ $$

Answered by prakash jain last updated on 02/Feb/15

y=(1+(x/n))^(nx)   ln y=nxln (1+(x/n))=((xln (1+(x/n)))/(1/n))  lim_(n→∞) ln y=lim_(n→∞) x∙(((−(x/n^2 ))/(1+(x/n)))/(−(1/n^2 )))=(x^2 /(1+(x/n)))=x^2   lim_(n→∞) f_n (x)=e^x^2

$${y}=\left(\mathrm{1}+\frac{{x}}{{n}}\right)^{{nx}} \\ $$$$\mathrm{ln}\:{y}={nx}\mathrm{ln}\:\left(\mathrm{1}+\frac{{x}}{{n}}\right)=\frac{{x}\mathrm{ln}\:\left(\mathrm{1}+\frac{{x}}{{n}}\right)}{\frac{\mathrm{1}}{{n}}} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}ln}\:{y}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}{x}\centerdot\frac{\frac{−\frac{{x}}{{n}^{\mathrm{2}} }}{\mathrm{1}+\frac{{x}}{{n}}}}{−\frac{\mathrm{1}}{{n}^{\mathrm{2}} }}=\frac{{x}^{\mathrm{2}} }{\mathrm{1}+\frac{{x}}{{n}}}={x}^{\mathrm{2}} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{f}_{{n}} \left({x}\right)={e}^{{x}^{\mathrm{2}} } \\ $$

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