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Question Number 50954 by afachri last updated on 22/Dec/18

$$\mathrm{i}\mathrm{was}\mathrm{evaluating}$$$$\underset{x\to \mathrm{\infty }}{\lim }\frac{x(\sqrt[x]{x^{}}-1)}{\log x}$$$$\mathrm{and}\mathrm{got}0\mathrm{as}\mathrm{the}\mathrm{product}.\mathrm{is}\mathrm{it}\mathrm{true},\mathrm{My}\mathrm{Fellows}??$$

Commented by Abdo msup. last updated on 23/Dec/18

$$letA\left(x\right)=\frac{x(x^{\frac{1}{x}}-1)}{ln\left(x\right)}changementln\left(x\right)=tgive$$$$A\left(x\right)=\frac{e^t({\left(e^t\right)}^{e^{-t}}-1)}{t}=\frac{e^t(e^{t{e}^{-t}}-1)}{t}$$$$x\to +\mathrm{\infty }\Rightarrow t\to +\mathrm{\infty }butA\left(x\right)=\frac{e^{(1+e^{-t})t}(1-e^{-t{e}^{-t}})}{t}$$$$e^{-t{e}^{-t}}\sim 1-t{e}^{-t}\Rightarrow 1-e^{-t{e}^{-t}}\sim t{e}^{-t}\Rightarrow$$$$A\left(x\right)\sim \frac{t{e}^{e^{-t}}}{t}\to 1(t\to +\mathrm{\infty })\Rightarrow {lim}_{x\to +\mathrm{\infty }}A\left(x\right)=1.$$

Commented by afachri last updated on 24/Dec/18

$$\mathrm{thnk}\mathrm{you}\mathrm{Sir}$$

Answered by Smail last updated on 22/Dec/18

$$\underset{x\to \mathrm{\infty }}{lim}\frac{x(e^{\frac{lnx}{x}}-1)}{lnx}=L$$$$lett=\frac{1}{x}$$$$L=\underset{t\to 0^{+}}{lim}\frac{e^{-tlnt}-1}{-tlnt}=\frac{1+(-tlnt)-1}{-tlnt}=1$$

Commented by afachri last updated on 22/Dec/18

$$\mathrm{Thanks}\mathrm{alot}\mathrm{for}\mathrm{answer}\mathrm{Mr}\mathrm{Smail}.\mathrm{i}\mathrm{will}$$$$\mathrm{figure}\mathrm{to}\mathrm{understand}\mathrm{it}.\mathrm{thank}\mathrm{you}\mathrm{very}$$$$\mathrm{much}.$$

Commented by afachri last updated on 22/Dec/18

$$\mathrm{Mr}\mathrm{Smail},\mathrm{why}\frac{1}{\log x}=\frac{1}{\ln x}????$$$$\mathrm{I}\mathrm{think}\mathrm{they}\mathrm{have}\mathrm{different}\mathrm{basic}\log$$$$\mathrm{number}.\mathrm{Would}\mathrm{you}\mathrm{be}\mathrm{kind}\mathrm{explain}$$$$\mathrm{to}\mathrm{me}????$$

Commented by Smail last updated on 23/Dec/18

$$Somepeopleuseloginsteadofln,soIthought$$$$thatyoudidthesamehere.$$$$log\left(x\right)={log}_{10}\left(x\right)=\frac{lnx}{ln\left(10\right)}$$$$Sotheanswerwouldbe$$$$\underset{t\to 0^{+}}{lim}\frac{e^{-tlnt}-1}{-t\frac{ln\left(t\right)}{ln\left(10\right)}}=\underset{t\to 0^{+}}{lim}\frac{1-tlnt-1}{-tlnt}\times ln\left(10\right)$$$$=ln\left(10\right)$$

Commented by afachri last updated on 23/Dec/18

$$\mathrm{Ok}\mathrm{thank}\mathrm{you}\mathrm{Mr}\mathrm{Smail}.:)$$

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