Question Number 5616 by sanusihammed last updated on 22/May/16
![Find [1 + 3n^(−1) ]^n Limit as n →∞ Please help.](https://www.tinkutara.com/question/Q5616.png)
$${Find}\: \\ $$$$ \\ $$$$\left[\mathrm{1}\:+\:\mathrm{3}{n}^{−\mathrm{1}} \right]^{{n}} \\ $$$$ \\ $$$${Limit}\:{as}\:{n}\:\rightarrow\infty \\ $$$$ \\ $$$$ \\ $$$${Please}\:{help}. \\ $$
Answered by FilupSmith last updated on 22/May/16
![L=lim_(n→∞) (1+(3/n))^n L=lim_(n→∞) exp(nln(1+(3/n))) L=lim_(n→∞) exp(((ln(1+(3/n)))/(1/n))) L′Hopital′s Law L=lim_(n→∞) exp((([((−6n^(−2) )/(1+3n^(−1) ))])/(−2n^(−2) ))) L=lim_(n→∞) exp((6/n^2 )×(1/(1+(3/n)))×(n^2 /2)) L=lim_(n→∞) exp(3×(1/(1+(3/n)))) L=lim_(n→∞) exp(3×(1/((n+3)/n))) L=lim_(n→∞) exp(3×(n/(n+3))) L′Hopital′s Law L=lim_(n→∞) exp(3×(1/1)) L=e^3 ∴lim_(n→∞) (1+(3/n))^n = e^3](https://www.tinkutara.com/question/Q5617.png)
$${L}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{1}+\frac{\mathrm{3}}{{n}}\right)^{{n}} \\ $$$${L}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{exp}\left({n}\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{3}}{{n}}\right)\right) \\ $$$${L}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{exp}\left(\frac{\mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{3}}{{n}}\right)}{\frac{\mathrm{1}}{{n}}}\right) \\ $$$${L}'{Hopital}'{s}\:{Law} \\ $$$${L}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{exp}\left(\frac{\left[\frac{−\mathrm{6}{n}^{−\mathrm{2}} }{\mathrm{1}+\mathrm{3}{n}^{−\mathrm{1}} }\right]}{−\mathrm{2}{n}^{−\mathrm{2}} }\right) \\ $$$${L}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{exp}\left(\frac{\mathrm{6}}{{n}^{\mathrm{2}} }×\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{3}}{{n}}}×\frac{{n}^{\mathrm{2}} }{\mathrm{2}}\right) \\ $$$${L}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{exp}\left(\mathrm{3}×\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{3}}{{n}}}\right) \\ $$$${L}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{exp}\left(\mathrm{3}×\frac{\mathrm{1}}{\frac{{n}+\mathrm{3}}{{n}}}\right) \\ $$$${L}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{exp}\left(\mathrm{3}×\frac{{n}}{{n}+\mathrm{3}}\right) \\ $$$${L}'{Hopital}'{s}\:{Law} \\ $$$${L}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{exp}\left(\mathrm{3}×\frac{\mathrm{1}}{\mathrm{1}}\right) \\ $$$${L}={e}^{\mathrm{3}} \\ $$$$ \\ $$$$\therefore\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{1}+\frac{\mathrm{3}}{{n}}\right)^{{n}} \:=\:{e}^{\mathrm{3}} \\ $$
Commented by sanusihammed last updated on 22/May/16
$${Thanks}.\:{i}\:{appreciate} \\ $$