$\lim_{x \to 0^+}(x |\ln x|)^\frac{1}{x}$

Question

I have to evaluate the following limit $$ \lim_{x \to 0^+}(x |\ln x|)^\frac{1}{x}$$ I tried with Hopital but I can't understand where I'm making mistakes. The final result is 0. Can someone help me?

• ok I see that hte function in not well defined for $x \rightarrow 0^+$, and supposing that the right function is

$$(x|\ln x|)^{\frac{1}{x}}$$

$$\lim_{x \to 0^+}(x |\ln x|)^\frac{1}{x}=\\ \lim_{x \to 0^+}e^{\log((x |\ln x|)^\frac{1}{x})} = \lim_{x \to 0^+}e^{\frac{\log((x |\ln x|))}{x}}= +\infty $$

Answer 1

Check this $$ \lim_{x \to 0^+}(x |\ln x|)^\frac{1}{x} $$ Take the logarithm of the expression $(x |\ln x|)^\frac{1}{x}$ and call it $L$ $$ L = \frac{1}{x}\ln(x |\ln x|) $$ Since $$ \lim_{x \rightarrow 0^+}(x |\ln x|) = 0 $$ that means $$ L = \frac{\ln(0^+)}{0^+} = -\infty $$ That means that the expression of which the logarithm is $L$ is $$ \lim_{x \to 0^+}(x |\ln x|)^\frac{1}{x} = 0 $$

Answer 2

As noticed in the comments, since $\log x<0$ the expression for the given limit is not well defined and we should consider

$$\lim_{x \to 0^+}(x |\ln x|)^\frac{1}{x}$$

With reference to this expression, firstly we need to show that

$$x |\ln x| \to 0$$

then note that $0^\infty$ is not an indeterminate form.

Answer 3

For $0<x<1/e$, $1< \log\big(\frac1x\big)$ and so $$ \begin{align} (x|\log x|)^{1/x}&=\exp\Big(\frac{1}{x}\big(\log x+ \log(|\log x|)\big)\Big)\\ &=\exp\Big(\frac{\log x +\log\big(\log\tfrac1x\big)}{x}\Big)\\ &=\exp\left(\frac{\log x}{x}\Big(1+ \frac{\log(\log\tfrac1x)}{\log x} \Big)\right)\xrightarrow{x\rightarrow0+}0 \end{align} $$

since $\frac{\log x}{x}\xrightarrow{x\rightarrow0+}-\infty$, and $\frac{\log\big(\log\tfrac1x\big)}{\log x}\stackrel{u=\log(1/x)}{=}-\frac{\log u}{u}\xrightarrow{u\rightarrow\infty}0$

Answer 4

In this problem letting $x\to 0^+$ along a certain sequence can lead to good insight. For example, you could let $x_n= 1/e^n.$ Then

$$(x_n|\ln x_n|)^{1/x_n} = (n/e^n)^{e^n}.$$

So think about this: That $n/e^n\to 0$ is well known at this level. Thus for $n$ large, $n/e^n<1/2.$ And what is $(1/2)^{e^n}?$ It's $1/2$ raised to a huge power. That suggests $0$ is the limit.