TITLE: Finding $\lim\limits_{x\to0}x^2\ln (x)$ without L'Hospital
QUESTION [9 upvotes]: I am preparing a resit for calculus and I encountered a limit problem.
The problem is the following: $\lim\limits_{x\to0}x^2\ln (x)$
I am not allowed to use L'Hospital.
Please help me, I am stuck for almost an hour now.

REPLY [2 votes]: For an elementary way consider this:
$\lim_{x\to 0} x^2 \ln x$
$\lim_{x\to 0} \ln {e^{x^2}} \ln x$
$\lim_{x\to 0} \ln [(e^{x^2})^{\ln x}]$
$\lim_{x\to 0} \ln[(e^{\ln x})^{x^2}]$
$\lim_{x\to 0} \ln x^{x^2}$
$\lim_{x\to 0} \ln (x^x)^x$
It is a well-known fact that $\lim_{x\to 0} x^x=1$ , so if we substitute that in we get $\lim_{x\to 0} \ln 1^x$ , or $\lim _{x\to 0} \ln 1=0$ . So our original limit is equal to $0$ . Let me know if anything is not clear. Hope this helps.