I want to evaluate this limit :$$\lim_{x\to 0}x\ln^2|x|$$

I wanted to use L'Hôpital's rule for this: $\lim\limits_{x\to0-}\frac{\ln^2|x|}{\tfrac{1}{x}}$, but I don't know how to differentiate the logarithm function, because of the absolute value.

My other question: do I have to calculate the one sided limits first?


Start by noting that $x\mapsto \text{abs}(x)\color{grey}{=|x|}$ is differentiable in $\mathbb R\setminus \{0\}$.

Then, for all $x\in \mathbb R\setminus \{0\}$, $$\begin{align} (\log\circ \text{abs})'(x)&=\log '(|x|)\cdot \text{abs}'(x)\\ &=\begin{cases} \log'(x)\cdot 1, &\text{if }x>0\\ \log'(-x)\cdot(-1), &\text{if }x<0\end{cases}\\ &=\begin{cases} \dfrac 1 x, &\text{if }x>0\\ \dfrac 1 x, &\text{if }x<0\end{cases}\\ &=\dfrac 1 x.\end{align}$$

As for the limit, one can guess it is zero just from looking as the function because one can look at $\log$ as a polynomial whose degree is infinitely closer to $0$, but not quite there, that is, the degree of the "polynomial $\log$" is $\varepsilon$, with $0<\varepsilon$ and $\forall y\in \mathbb R(y>0\implies \varepsilon <y)$.

This non-sense characterization actually works, immediately yielding $\lim \limits_{x\to 0}\left(x^\alpha(\log(|x|)^\beta\right)=0$, for all $\alpha, \beta>0$.


It seems, you want to evaluate that for $x \in \mathbb{R}$.
I recommend interpreting the absolute value as a piecewise defined function

$$\left| x \right| = \begin{cases} x &,x\leq0\\-x&,x<0\end{cases} .$$

This than answers both questions as it leads to the answer to the first one and includes the second: The limit only exists if it equals for both sides.


You can compute it directly by this substitution: $$\lim_{x\to0} x\ln^2|x|=\lim_{y\to\infty}\pm e^{-y}\ln^2(e^{-y})=\lim_{y\to\infty}\pm\frac{y^2}{e^y}=0$$

  • $\begingroup$ This is a nice deal - but I think one should be careful when using always. $\endgroup$ – Bastian Ebeling Apr 23 '14 at 9:49
  • 1
    $\begingroup$ @BastianEbeling I meant limits of the type $x^\alpha ln^\beta x$, but I agree this wording is confusing, so I removed it. $\endgroup$ – user2345215 Apr 23 '14 at 10:42

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