How do I prove that $\lim_{x\to0^+} x\cdot\ln x=0$ How do I prove (without L'Hôpital's rule) that  $$\lim_{x\to0^+} x\cdot\ln x = 0$$.
I'm trying to get some intuitive sense for this, but it's quite hard. It's like trying to prove that $x$ goes faster to $0$ then $\ln x$ goes to $-\infty$, right ?
I tried this, for $x\in(0,1)$
$$x \ln x=x⋅-\int_x^1 \frac{1}{t}dt>x\cdot\frac{(1-x)}{-x}=-1+x$$
So when $x\to0^+$, I conclude that $x\cdot\ln x\ge-1$.
 A: 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: $\ln x < \ln 2 + \sqrt{x} =o(x)$.)
A: $$x \ln x=-x⋅\int_x^1 \frac{1}{t}dt$$
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....
A: 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.
A: Another method is to use Taylor's series: $\ln(1-y)=-y-\frac{y^2}{2}-\frac{y^3}{3}-\ldots$
$$\therefore \lim_{x\to 0}x\cdot \ln x$$
$$=\lim_{x\to 0}x\cdot \lim_{x\to 0}\ln x$$
$$=\lim_{x\to 0}x\cdot \lim_{x\to 0}\ln (1-(1-x))$$
$$=\lim_{x\to 0}x\cdot \lim_{x\to 0}\left(-(1-x)-\frac{(1-x)^2}{2}-\frac{(1-x)^3}{3}-\ldots\right)$$
$$=0\cdot \left(-1-\frac{1}{2}-\frac{1}{3}-\ldots\right)$$
$$=0$$
Proved.
