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Question Number 52208 by SUJIT420 last updated on 04/Jan/19

Commented by maxmathsup by imad last updated on 04/Jan/19

$$itsnotcitse.$$

Commented by maxmathsup by imad last updated on 04/Jan/19

$$letU\left(x\right)={(1+x)}^{\frac{1}{x}}-eandV\left(x\right)=xwehave$$$$U\left(x\right)=e^{\frac{1}{x}ln(1+x)}-e\Rightarrow {lim}_{x\to 0}U\left(x\right)=0hospitaltheoremgive$$$${lim}_{x\to 0}\frac{U\left(x\right)}{x}={lim}_{x\to 0}{U}^{{}^{\prime }}\left(x\right)but{U}^{{}^{\prime }}\left(x\right)={\left(\frac{ln(1+x)}{x}\right)}^{{}^{\prime }}U\left(x\right)$$$$=\frac{\frac{x}{1+x}-ln(1+x)}{x^2}U\left(x\right)=\frac{x-(1+x)ln(1+x)}{x^2}U\left(x\right)but$$$$ln(1+x)\sim x\Rightarrow x-(1+x)ln(1+x)\sim x-x(1+x)=-x^2\Rightarrow U^{{}^{\prime }}\left(x\right)\sim -U\left(x\right)(x\to 0)$$$$wehave{ln}^{{}^{\prime }}(1+x)=\frac{1}{1+x}=1-x+o\left(x^2\right)\Rightarrow ln(1+x)=x-\frac{x^2}{2}+o\left(x^3\right)\Rightarrow$$$$\frac{ln(1+x)}{x}=1-\frac{x}{2}+o\left(x^2\right)\Rightarrow e^{\frac{ln(1+x)}{x}}=e^{1-\frac{x}{2}+o\left(x^2\right)}=e(1-\frac{x}{2}+o\left(x^2\right))\Rightarrow$$$$e^{\frac{ln(1+x)}{x}}-e=-\frac{ex}{2}+o\left(x^2\right)\Rightarrow \frac{U\left(x\right)}{x}=-\frac{e}{2}+o\left(x\right)\Rightarrow {lim}_{x\to 0}\frac{U\left(x\right)}{x}=-\frac{e}{2}$$$$andweseethathospitaltheoremdontworkgoodinthiscase.$$

Commented by afachri last updated on 05/Jan/19

$$\mathrm{is}\mathrm{it}\mathrm{possible}\mathrm{to}\mathrm{transform}{(1+x)}^{\frac{1}{x}}-e\mathrm{into}\mathrm{Taylor}$$$$\mathrm{serie}\mathrm{Sir}???\mathrm{because}\mathrm{i}\mathrm{got}\mathrm{stuck}\mathrm{to}\mathrm{find}\mathrm{the}$$$$\mathrm{derrivative}\mathrm{of}{(1+x)}^{\frac{1}{x}}$$

Answered by Smail last updated on 05/Jan/19

$$\underset{x\to 0}{lim}\frac{{(1+x)}^{1/x}-e}{x}=\underset{x\to 0}{lim}\frac{e^{\frac{ln(1+x)}{x}}-e}{x}$$$$ln(1+x)\approx x-\frac{x^2}{2}near0$$$$so\underset{x\to 0}{lim}\frac{{(1+x)}^{1/x}-e}{x}=\underset{x\to 0}{lim}\frac{e^{\frac{x-\frac{x^2}{2}}{x}}-e}{x}$$$$=\underset{x\to 0}{lim}\frac{e^{1-\frac{x}{2}}-e}{x}=\underset{x\to 0}{lim}\frac{e(e^{-\frac{x}{2}}-1)}{x}$$$$e^t\approx 1+tnear0$$$$\underset{x\to 0}{lim}\frac{{(1+x)}^{1/x}-e}{x}=\underset{x\to 0}{lim}\frac{e(1-\frac{x}{2}-1)}{x}=\underset{x\to 0}{lim}\frac{-xe}{2x}$$$$=\frac{-e}{2}$$

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