Limit of a natural log

Question

I have this limit

$\lim_{x\to0^+}\big(\dfrac1x\big)^{\ln(x)}$

Can somebody help me understand how this limit equals $0$?

Answer 1

In these cases it's useful to look at the logarithm of your function, that is $$ \ln\left(\frac{1}{x}\right)^{\!\ln x}=(\ln x)\ln\frac{1}{x}=-(\ln x)^2 $$ Therefore you have to compute $$ \lim_{x\to0^+}e^{-(\ln x)^2} $$ Can you go on?

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A word on explanation: if you denote by $y$ your given expression and $z=\ln y$, then $y=e^z$. In this case it turns out that $z$ can be easily simplified and $y$ rewritten in a simpler way for computing the limit.

In many other cases, when you have a limit in the form $$ \lim f(x)^{g(x)} $$ the limit of $z=\ln(f(x)^{g(x)}=g(x)\ln f(x)$ can be computed more easily than the limit of $y=f(x)^{g(x)}$ and then we can use the continuity of the exponential. Namely, if $\lim z=a$, then $$ \lim y = \lim e^z = e^a $$ with almost obvious extensions when $a=\infty$ or $a=-\infty$.

Answer 2

$$\underset{x\to 0^+}{\text{lim}}\left(\frac{1}{x}\right)^{\ln (x)}=\\\underset{x\to 0^+}{\text{lim}}x^{-(1)\ln (x)}=\\\underset{x\to 0^+}{\text{lim}}x^{-\ln (x)}=$$ Using property:

$$a^b=e^{b \ln (a)}$$

and then:

$$\\\underset{x\to 0^+}{\text{lim}}e^{-\ln (x) \ln (x)}=\\\underset{x\to 0^+}{\text{lim}}e^{-\ln ^2(x)}=0$$