Find the limit of $x \ln[\ln(x^{-2})]$ as $x$ tends to $0$

Question

$$\lim_{x\to 0}x \ln\left( \ln\left(\frac{1}{x^2}\right)\right)$$

What's the proper way to evaluate the limit of such a function ?

Answer 1

One time l'Hopital (H): \begin{align} \lim_{x\to 0}x\ln\left(\ln\left(\frac{1}{x^2}\right) \right)=\lim_{x\to 0}\frac{\ln(-\ln(x^2))}{1/x}\stackrel{H}{=}\lim_{x\to 0}\frac{-2x}{\ln(x^2)} \end{align} Now you can finish right?

Answer 2

then you will have $$\lim_{x \to 0}x\ln(-2\ln|x|)=0$$

Answer 3

Let $\frac1{x^2}=e^{2y}$ with $y\to +\infty$

$$\lim_{x\to 0}x \ln\left( \ln\left(\frac{1}{x^2}\right)\right)=\lim_{y\to +\infty}\frac1{e^y} \ln(2y)=\lim_{y\to +\infty}\frac{2y}{e^y} \frac{\ln(2y)}{2y}=0$$