I was trying to compute the following:
$$\lim_{x\to 0+}\frac{\ln x}{x^2}$$
And it stumped me after doing the following work:
Applied the product rule for limits:
$$\lim_{x\to 0+}\frac{\ln x}{x^2} = \lim_{x\to 0+}\frac{1}{x^2}\bullet\lim_{x\to 0+} \ln x$$
$$\lim_{x\to 0+} \ln x = -\infty$$
$$\lim_{x\to 0+}\frac{\ln x}{x^2}= \lim_{x\to 0+}\frac{1}{x^2} \bullet -\infty$$
Now, for as to what to do here, I was stumped, and WolframAlpha did the following:
$$\lim_{x\to 0+}\frac{1}{x^2} = \lim_{x\to 0+}e^{-2\ln x}$$
And then, to my confusion, did this:
$$\lim_{x\to 0+}\frac{1}{x^2} = e^{\lim_{x\to 0+}-2\ln x}$$
And with that, proceeded to solve:
$$\lim_{x\to 0+}\frac{1}{x^2} = e^{-2\ \bullet \ -\infty}$$
$$\lim_{x\to 0+}\frac{1}{x^2} = \infty$$
For the final result:
$$\infty \bullet -\infty = -\infty$$
What I don't understand, however, is how
$$\lim_{x\to 0+}e^{-2\ln x} = e^{\lim_{x\to 0+}-2\ln x}$$
preserves equality. How is this justifiable? Is this always applicable?
$\lim\limits_{x\rightarrow0}\frac{\ln{x}}{x^2}$ is absurd, but $$\lim\limits_{x\rightarrow0^+}\frac{\ln{x}}{x^2}=-\infty$$ because for $x\rightarrow0^+$ we have $\ln{x}\rightarrow-\infty$ by definition of $\ln$ and $\frac{1}{x^2}\rightarrow+\infty$.
Now we can use the definition of $\lim$ if you wish.