Evaluate $\lim_{x\to 0} \frac{\log(x^2)}{1/x}$

Question

$\displaystyle \lim_{x\to 0} \frac{\log(x^2)}{1/x}=0$ (where the result is 0 because from hierarchy of infinities we know that $1/x$ goes to infinity faster than $\log(x^2)$)

Can I solve the limit in this way without using de l'hospital? To do: tell me if my idea is right.

Answer 1

You didn't want to use L'Hopital's rule, but we can use it to check:

The top one's derivative is $\frac{2}{x}$ (assuming your $\log$ is the natural log, but it works either way because change of base adds a multiplicative constant)

The bottom one's derivative is $-\frac{1}{x^2}$.

$\frac{2}{x} * -x^2$ = $-2x$.

And we can plug in $x = 0$ to get $0$. So yes, it is correct!

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Note that your "hierarchy of infinities" directly follows from the essence of L'Hopital's rule, so you're essentially reinventing the wheel here.

Answer 2

But why is the hierarchy ordered as stated? I think that's a proof in the wrong direction. With $y:=-\ln x$, note$$\lim_{x\to0^+}2x\ln x=2\lim_{y\to\infty}ye^{-y}=0$$because $\int_0^\infty ye^{-y}dy=1$ is a finite integral with continuous non-negative integrand. Then use that $x\ln(x^2)$ is odd.