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Question Number 6944 by Tawakalitu. last updated on 03/Aug/16

Prove that:    e^x = Limit(u → + ∞) (1 + (x/u))^u

$${Prove}\:{that}: \\ $$$$ \\ $$$${e}^{{x}} =\:{Limit}\left({u}\:\rightarrow\:+\:\infty\right)\:\left(\mathrm{1}\:+\:\frac{{x}}{{u}}\right)^{{u}} \: \\ $$

Answered by sou1618 last updated on 03/Aug/16

L=lim_(u→+∞) (1+(x/u))^u   lim_(p→+∞) (1+(1/p))^p =e  now...  p=(u/x)  L=lim_(u→+∞) (1+(x/u))^((u/x)×x)   L={lim_(u→+∞) (1+(x/u))^(u/x) }^x   L=e^x

$${L}=\underset{{u}\rightarrow+\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{{x}}{{u}}\right)^{{u}} \\ $$$$\underset{{p}\rightarrow+\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{1}}{{p}}\right)^{{p}} ={e} \\ $$$${now}...\:\:{p}=\frac{{u}}{{x}} \\ $$$${L}=\underset{{u}\rightarrow+\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{{x}}{{u}}\right)^{\frac{{u}}{{x}}×{x}} \\ $$$${L}=\left\{\underset{{u}\rightarrow+\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{{x}}{{u}}\right)^{\frac{{u}}{{x}}} \right\}^{{x}} \\ $$$${L}={e}^{{x}} \\ $$$$ \\ $$

Commented by Tawakalitu. last updated on 03/Aug/16

Thank you very much

$${Thank}\:{you}\:{very}\:{much} \\ $$

Answered by FilupSmith last updated on 03/Aug/16

Prove      e^x =lim_(u→∞) (1+(x/u))^u   L=lim_(u→∞) (1+(x/u))^u   L=lim_(u→∞)  e^(uln(1+(x/u)))   L=lim_(u→∞)  exp(((ln(1+(x/u)))/(1/u)))  L′Hopital′s Law  L=lim_(u→∞)  exp(((−(x/(u^2 +ux)))/(−(1/u^2 ))))  L=lim_(u→∞)  exp(((ux)/(u+x)))  Apply law again  L=lim_(u→∞)  exp((x/1))  ∴ L=e^x

$$\mathrm{Prove}\:\:\:\:\:\:{e}^{{x}} =\underset{{u}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{{x}}{{u}}\right)^{{u}} \\ $$$${L}=\underset{{u}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{{x}}{{u}}\right)^{{u}} \\ $$$${L}=\underset{{u}\rightarrow\infty} {\mathrm{lim}}\:{e}^{{u}\mathrm{ln}\left(\mathrm{1}+\frac{{x}}{{u}}\right)} \\ $$$${L}=\underset{{u}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{exp}\left(\frac{\mathrm{ln}\left(\mathrm{1}+\frac{{x}}{{u}}\right)}{\frac{\mathrm{1}}{{u}}}\right) \\ $$$${L}'{Hopital}'{s}\:{Law} \\ $$$${L}=\underset{{u}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{exp}\left(\frac{−\frac{{x}}{{u}^{\mathrm{2}} +{ux}}}{−\frac{\mathrm{1}}{{u}^{\mathrm{2}} }}\right) \\ $$$${L}=\underset{{u}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{exp}\left(\frac{{ux}}{{u}+{x}}\right) \\ $$$${Apply}\:{law}\:{again} \\ $$$${L}=\underset{{u}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{exp}\left(\frac{{x}}{\mathrm{1}}\right) \\ $$$$\therefore\:{L}={e}^{{x}} \\ $$

Commented by Tawakalitu. last updated on 03/Aug/16

Thanks so much

$${Thanks}\:{so}\:{much} \\ $$

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