limit-x-1-1-x-x-2-

Question Number 2478 by Syaka last updated on 21/Nov/15

$${limit}\:{x}\:\rightarrow\:\infty \\ $$$$\left(\mathrm{1}\:+\:\frac{\mathrm{1}}{{x}}\right)^{{x}\:+\:\mathrm{2}} \:=\:? \\ $$

Commented by John_Haha last updated on 21/Nov/15

$${e} \\ $$

Answered by Yozzi last updated on 21/Nov/15

L=lim_(x→∞) (1+(1/x))^(x+2) =lim_(x→∞) (1+(1/x))^2 (1+(1/x))^x L=(lim_(x→∞) (1+(1/x))^2 )(lim_(x→∞) (1+(1/x))^x ) L=((1+(1/∞))^2 )(e) L=1^2 ×e=e Proof for lim_(x→∞) (1+1/x)^x =e Let l=lim_(x→∞) (1+1/x)^x (∗). (1+1/x)^x >0 ∀x>0 so if l exists, l>0. Taking logs to base e on both sides of (∗). lnl=lim_(x→∞) ln(1+1/x)^x lnl=lim_(x→∞) xln(1+(1/x)) Let u=1/x⇒as x→∞,u→0. ∴lnl=lim_(u→0) (1/u)ln(1+u) Using the Maclaurin expansion for ln(1+u)=u−(u^2 /2)+(u^3 /3)−(u^4 /4)+(u^5 /5)−... −1<u≤1 ⇒lnl=lim_(u→0) (1/u)(u−(u^2 /2)+(u^3 /3)−(u^4 /4)+(u^5 /5)−...) lnl=lim_(u→0) (1−(u/2)+(u^2 /3)−(u^3 /4)+(u^4 /5)−...) lnl=1−0+0−0+0−... lnl=1⇒l=e ∴ lim_(x→∞) (1+(1/x))^x =e

$${L}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}+\mathrm{2}} =\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}} \\ $$$${L}=\left(\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} \right)\left(\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}} \right) \\ $$$${L}=\left(\left(\mathrm{1}+\frac{\mathrm{1}}{\infty}\right)^{\mathrm{2}} \right)\left({e}\right) \\ $$$${L}=\mathrm{1}^{\mathrm{2}} ×{e}={e} \\ $$$$ \\ $$$${Proof}\:{for}\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\mathrm{1}/{x}\right)^{{x}} ={e} \\ $$$${Let}\:{l}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\mathrm{1}/{x}\right)^{{x}} \:\:\left(\ast\right). \\ $$$$\left(\mathrm{1}+\mathrm{1}/{x}\right)^{{x}} >\mathrm{0}\:\forall{x}>\mathrm{0}\:{so}\:{if}\:{l}\:{exists},\:{l}>\mathrm{0}. \\ $$$${Taking}\:{logs}\:{to}\:{base}\:{e}\:{on}\:{both}\:{sides}\:{of}\:\left(\ast\right). \\ $$$${lnl}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}{ln}\left(\mathrm{1}+\mathrm{1}/{x}\right)^{{x}} \\ $$$${lnl}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}{xln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right) \\ $$$${Let}\:{u}=\mathrm{1}/{x}\Rightarrow{as}\:{x}\rightarrow\infty,{u}\rightarrow\mathrm{0}. \\ $$$$\therefore{lnl}=\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{{u}}{ln}\left(\mathrm{1}+{u}\right) \\ $$$${Using}\:{the}\:{Maclaurin}\:{expansion}\:{for} \\ $$$${ln}\left(\mathrm{1}+{u}\right)={u}−\frac{{u}^{\mathrm{2}} }{\mathrm{2}}+\frac{{u}^{\mathrm{3}} }{\mathrm{3}}−\frac{{u}^{\mathrm{4}} }{\mathrm{4}}+\frac{{u}^{\mathrm{5}} }{\mathrm{5}}−…\:\:\:−\mathrm{1}<{u}\leqslant\mathrm{1} \\ $$$$\Rightarrow{lnl}=\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{{u}}\left({u}−\frac{{u}^{\mathrm{2}} }{\mathrm{2}}+\frac{{u}^{\mathrm{3}} }{\mathrm{3}}−\frac{{u}^{\mathrm{4}} }{\mathrm{4}}+\frac{{u}^{\mathrm{5}} }{\mathrm{5}}−…\right) \\ $$$${lnl}=\underset{{u}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}−\frac{{u}}{\mathrm{2}}+\frac{{u}^{\mathrm{2}} }{\mathrm{3}}−\frac{{u}^{\mathrm{3}} }{\mathrm{4}}+\frac{{u}^{\mathrm{4}} }{\mathrm{5}}−…\right) \\ $$$${lnl}=\mathrm{1}−\mathrm{0}+\mathrm{0}−\mathrm{0}+\mathrm{0}−… \\ $$$${lnl}=\mathrm{1}\Rightarrow{l}={e} \\ $$$$\therefore\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}} ={e} \\ $$$$ \\ $$$$ \\ $$

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