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Question Number 8839 by tawakalitu last updated on 31/Oct/16

Prove that. lim_(n→∞) (1 + n)^(1/n) = e

$$\mathrm{Prove}\:\mathrm{that}.\: \\ $$$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{1}\:+\:\mathrm{n}\right)^{\mathrm{1}/\mathrm{n}} \:=\:\mathrm{e} \\ $$

Answered by FilupSmith last updated on 31/Oct/16

L=lim_(n→∞) (1 + n)^(1/n) L=lim_(n→∞) exp{(1/n)ln(1 + n)} L=exp{lim_(n→∞) (1/n)ln(1 + n)} L′Hopitals law L=exp{lim_(n→∞) (1/(n+1))} L=e^0 L=1 lim_(n→∞) (1 + n)^(1/n) =1 lim_(n→∞) (1 + n)^(1/n) ≠e

$${L}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{1}\:+\:{n}\right)^{\mathrm{1}/{n}} \\ $$$${L}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{exp}\left\{\frac{\mathrm{1}}{{n}}\mathrm{ln}\left(\mathrm{1}\:+\:{n}\right)\right\} \\ $$$${L}=\mathrm{exp}\left\{\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{n}}\mathrm{ln}\left(\mathrm{1}\:+\:{n}\right)\right\} \\ $$$$\mathrm{L}'\mathrm{Hopitals}\:\mathrm{law} \\ $$$${L}=\mathrm{exp}\left\{\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{{n}+\mathrm{1}}\right\} \\ $$$${L}={e}^{\mathrm{0}} \\ $$$${L}=\mathrm{1} \\ $$$$ \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{1}\:+\:{n}\right)^{\mathrm{1}/{n}} =\mathrm{1} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{1}\:+\:{n}\right)^{\mathrm{1}/{n}} \neq{e} \\ $$

Commented by tawakalitu last updated on 31/Oct/16

I really appreciate sir.

$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{sir}. \\ $$

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