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

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.

• Hint: use base $e$ to represent the exponential – tommy1996q Jan 3 at 20:17
• @tommy1996q do you want it to be like: $e^{ln(x)/x}*x-x$ – Yay Jan 3 at 20:19

$$\lim\limits_{x\to+\infty}x(x^{\frac{1}{x}}-1)=\lim\limits_{x\to+\infty}\dfrac{x^{\frac{1}{x}}-1}{\frac{1}{x}}=\lim\limits_{x\to+\infty}\dfrac{\frac{1-\ln{x}}{x^2}}{-\frac{1}{x^2}}=\lim\limits_{x\to+\infty}(\ln{x}-1)=+\infty$$

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

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

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

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

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 $$\lim_{t\to0^+}-t^t(1+\log t)=\infty$$

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

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

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

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

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

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