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Evaluate $$\lim_{x\to\infty} x^{1/x}\cdot x-x.$$

I looked at the graph on Desmos and it appears to start approaching infinity but at around 3 nonnilion it goes to 0. I know this is a inf-inf situation and do not know how to solve it. I think the best way to start would be to multiply by $x/x$ and than use l’hospitals rule to proceed.

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  • $\begingroup$ Please add your steps. Also please add suggestions on how to improve this post in the comment section and please make helpful edits. $\endgroup$ – Yay Jan 3 at 20:11
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    $\begingroup$ Hint: use base $e$ to represent the exponential $\endgroup$ – tommy1996q Jan 3 at 20:17
  • $\begingroup$ @tommy1996q do you want it to be like: $e^{ln(x)/x}*x-x$ $\endgroup$ – Yay Jan 3 at 20:19
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$\lim\limits_{x\to+\infty}x(x^{\frac{1}{x}}-1)=\lim\limits_{x\to+\infty}\dfrac{x^{\frac{1}{x}}-1}{\frac{1}{x}}=\lim\limits_{x\to+\infty}\dfrac{\frac{1-\ln{x}}{x^2}}{-\frac{1}{x^2}}=\lim\limits_{x\to+\infty}(\ln{x}-1)=+\infty$

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Note that $x(x^{1/x}-1)=x\left(e^{\frac1x\log(x)}-1\right)$. Moreover, since $e^x\ge 1+x$ we see that

$$\log(x)\le x\left(e^{\frac1x\log(x)}-1\right)$$

whence applying the squeeze theorem yields the coveted result

$$\lim_{x\to \infty}x(x^{1/x}-1)=\infty$$

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$$x(x^{\frac{1}{x}}-1)=x(e^{\frac{1}{x}log(x)}-1)$$

We have $$e^{\frac{1}{x}log(x)}=1+\frac{1}{x}log(x)+O(\frac{1}{x^2}log(x)^2)$$

$$x(e^{\frac{1}{x}log(x)}-1)=log(x)+O(\frac{1}{x}log(x)^2)$$ which tends to infinity when $x$ tends to infinity

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If you transform $x=1/t$, the limit becomes $$ \lim_{t\to0^+}\frac{1}{t^t}\frac{1-t^t}{t} $$ The first fraction has limit $1$, so we just need to evaluate the second one: the derivative of $f(t)=t^t=e^{t\log t}$ is $f'(t)=t^t(1+\log t)$, so we remain with $$ \lim_{t\to0^+}-t^t(1+\log t)=\infty $$

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$$\lim_{x\to\infty} x^{1/x}\cdot x-x= \lim _{x\to \infty}(x(x^{\frac{1}{x}}-1))=\lim _{x\to \infty}\left(\frac{-1+x^{\frac{1}{x}}}{\frac{1}{x}}\right)$$ If I apply the Hospital's theorem we have:

$$=\lim _{x\to \infty}\left(\frac{x^{\frac{-2x+1}{x}}\left(-\ln \left(x\right)+1\right)}{-\frac{1}{x^2}}\right)=\lim_{x\to \infty}\frac{x^{\frac{-2x+1}{x}}\left(-\ln \left(x\right)+1\right)}{\frac{1}{x^2}}=\lim _{x\to \infty}\frac{x^{\frac{-2x+1}{x}}\left(-\ln \left(x\right)+1\right)x^2}{1}$$ i.e. $$=\lim _{x\to \infty}-x^2x^{\frac{-2x+1}{x}}\left(-\ln \left(x\right)+1\right)=\lim _{x\to \infty}\left(-x^{\frac{-2x+1}{x}+2}\left(-\ln \left(x\right)+1\right)\right)$$ $$=-\lim _{x\to \infty }\left(x^{\frac{-2x+1}{x}+2}\right)\cdot \lim _{x\to \infty}\left(-\ln \left(x\right)+1\right)=-1\cdot (-\infty)=\infty.$$

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Set $x:=e^y$, and let $y \rightarrow \infty$.

$x^{1/x}=(e^y)^{e^{-y}}=e^{ye^{-y}}$;

$x(x^{1/x}-1)=$

$e^y(e^{ye^{-y}}-1)\ge$

$e^y(1+ye^{-y}-1)=y.$

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