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?
Best Answer
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: $$\begin{align*} \lim_{x\to 0^+}x(\ln x)^2 &= -2\lim_{x\to 0^+}x\ln(x) = -2\lim_{x\to 0^+}\frac{\ln(x)}{x^{-1}}\\ &= -2\lim_{x\to 0^+}\frac{x^{-1}}{-x^{-2}} = -2\lim_{x\to 0^+}(-x) = (-2)(0) = 0. \end{align*}$$