How do I prove that $\lim_{x\to0^+} x\cdot\ln x=0$

$$x \ln x=-x⋅\int_x^1 \frac{1}{t}dt$$

Now, let $0<a$ be arbitrary. Then, for all $x \in (0,1)$ we have

$$\frac{1}{t^{1-a}}\leq \frac{1}{t} \leq \frac{1}{t^{1+a}}$$

Thus

$$-x⋅\int_x^1 \frac{1}{t^{1+a}}dt \leq -x \ln(x) \leq -x⋅\int_x^1 \frac{1}{t^{1-a}}dt$$ $$-x⋅\frac{x^a-1}{a} \leq -x \ln(x) \leq -x⋅\frac{x^{-a}-1}{-a}⋅$$

now let $x \to 0$.

P.S. The argument works directly with any $0<a<1$, so picking $a=\frac{1}{2}$ makes the proof much cleaner....

Since $\ln(\frac{1}{x})=-\ln x$, $$\lim_{x\to 0^{+}} x \cdot \ln x = \lim_{x\to \infty} \frac{1}{x} \cdot\ln \frac{1}{x} = - \lim_{x\to \infty} \frac{\ln x}{x}$$

But $\ln x$ is "slower" than $x$ (since $e^x$ is faster than $x$), so this limit is zero.

(Rigorously: for positive $x$, $2 e^x > x^2$ (by comparing power series of both sides). Plug $\sqrt{x}$ instead of $x$ and take logarithm of both sides: $\ln x < \ln 2 + \sqrt{x} =o(x)$.)

Make the change $x = e^{-y}$: $$ \lim_{x \to 0} x \ln x = - \lim_{y \to \infty} y e^{-y} $$

Show that for $y > 0$: $$ e^y > \frac{y^2}{2!} $$

And use the squeeze theorem.