calculus - $lim_{xto0^{+}} x ln x$ without l'Hopital's rule

I have a midterm coming up and on the past exams the hard question(s) usually involve some form of $\lim_{x\to0^{+}} x \ln x$. However, we're not allowed to use l'Hopital's rule, on this year's exam anyways.

So how can I evaluate said limit without l'Hopital's rule? I got somewhere with another approach, don't know if it's useful:

1. $\lim_{x\to0^{+}} x \ln x = \lim_{x\to0^{+}} x^2 \ln (x^2) = L$


2. $= (\lim_{x\to0^{+}} 2x)(\lim_{x\to0^{+}} x \ln x)$


3. $= 0 * L$

Then I just need to prove that L is finite/exists (which means it must be 0)

Answer

The idea you described is a very nice one. We fill in the details.

We consider, as in the OP, $x^2\ln(x^2)$, that is, $(2x)(x\ln x)$. If we can show that $x\ln x$ is bounded
near $0$, it will follow by Squeezing that $\displaystyle\lim_{x\to 0} x^2\ln(x^2)=0$, and therefore $\displaystyle\lim_{t\to 0^+}t\ln t=0$.

Let $f(x)=x\ln x$. Then $f'(x)=1+\ln x$. It follows that $f(x)$ is decreasing in the interval $(0,e^{-1})$. It reaches a minimum value of $-e^{-1}$ at $x=e^{-1}$.

Since $f(x)$ is negative in our interval, we have $|x\ln x|\le e^{-1}$ in the interval, and we have shown boundedness.