Find $\lim_{x\to0^+}\ x\left(\ln x\right)^{2}$ Q:

Find $\lim_{x\to0^+}\ x\left(\ln x\right)^{2}$

My approach:

$$\lim_{x\to0^+}\ x\cdot\left(\ln x\right)^{2}\ \to0\cdot\infty$$
$$\lim_{x\to0^+\ }\left(\frac{x}{\left(\ln x\right)^{-2}}\right)\to\frac{0}{0}$$
Applying LH Rule, $$\lim_{x\to0^+}\left(\frac{x}{-2\ln\left(x\right)^{-3}}\right)\to\frac{0}{0}$$

but this just goes on..the indeterminacy always remain. how do I calculate the limit? Where am I going wrong? I do not know anything about transformations and stuff, I am in high school, is there any other way to find this?
 A: Try instead the other rewriting:
$$\begin{align*}
\lim_{x\to 0^+}x(\ln(x))^2 &= \lim_{x\to 0^+}\frac{(\ln x)^2}{x^{-1}}
= \lim_{x\to 0^+}\frac{2\ln(x)/x}{-x^{-2}}\\
&= \lim_{x\to 0^+}-2x\ln(x) = -2\lim_{x\to 0^+}x\ln(x).
\end{align*}$$
This is still an indeterminate, but it looks easier than the original one. So we do it again:
$$\begin{align*}
\lim_{x\to 0^+}x(\ln x)^2 &= -2\lim_{x\to 0^+}x\ln(x) = -2\lim_{x\to 0^+}\frac{\ln(x)}{x^{-1}}\\
&= -2\lim_{x\to 0^+}\frac{x^{-1}}{-x^{-2}} = -2\lim_{x\to 0^+}(-x) = (-2)(0) = 0.
\end{align*}$$
A: When assigning one part to the numerator and the other part to the denominator results in a sequence of l'Hospital's rule applications that get gradually worse, swap which is which.
$$  \lim_{x \rightarrow 0^+} x \cdot (\ln x)^2 = \lim_{x \rightarrow 0^+} \frac{(\ln x)^2}{1/x}  \text{.}  $$
This gives an infinite version of l'Hospital's rule.
$$  \cdots = \lim_{x \rightarrow 0^+} \frac{2(\ln x) \cdot \frac{1}{x}}{-1/x^2} = \lim_{x \rightarrow 0^+} \frac{2(\ln x)}{-1/x}  \text{,}  $$
and with swapping, we see that applications of the rule are incremental improvements.  Just one more application and the logarithm vanishes...
A: If we take as given that $\lim_{u\to0^+}u\ln u=0$, then, letting $x=u^2$, we have
$$\lim_{x\to0^+}x(\ln x)^2=\lim_{u\to0^+}u^2(\ln u^2)^2=(2\lim_{u\to0^+}u\ln u)^2=(2\cdot0)^2=0$$
There are various ways of establishing $\lim_{u\to0^+}u\ln u=0$, including L'Hopital:
$$\lim_{u\to0^+}u\ln u=\lim_{u\to0^+}{\ln u\over1/u}=\lim_{u\to0^+}{1/u\over-1/u^2}=-\lim_{u\to0^+}u=0$$
and the definition of the natural logarithm as an integral:
$$|u\ln u|=\left|u\int_1^u{dt\over t}\right|=u\left(\int_u^{\sqrt u}{dt\over t} +\int_\sqrt u^1{dt\over t}\right)\le u\left({\sqrt u-u\over u} +{1-\sqrt u\over\sqrt u}\right)=2(\sqrt u-u)\to0$$
(where the inequality requires $u\le1$).
A: Let $f(x)=x\ln(x)^2$ then $f'(x)=\underbrace{\ln(x)}_{<0}\underbrace{(\ln(x)+2)}_{<0}>0$ in a neighborhood of $0+$.
So $f\nearrow$ near $0+$ and $f$ continuous on $(0,1]$ and $f\ge 0$ therefore $f$ is bounded on $[0,1]$.
But then $f(x)=(\sqrt{x})^2\ln((\sqrt{x})^2)^2=(\sqrt{x})^2\times 4\ln(\sqrt{x})^2=4\underbrace{f(\sqrt{x})}_{<M}\underbrace{\sqrt{x}}_{\to 0}\to 0$
Note that this proof is easily adaptable to $x\ln(x)^p$.
