Can you use L'Hospital for a "$0/\infty$" form as in the case $\lim_{x\to 0} \log(x)/\csc(x)$

Question

The original sum was to evaluate $$ \lim_{x\to 0} |x|^{\sin(x)}.$$ I converted it into $$\lim_{x\to 0} \mathrm{e}^{\sin(x)\log|x|}, $$ and then made $\sin x$ as $\csc x$ and shifted to denominator. My friend tells me to use L'Hospital's rule, but as far as I know its used for $0/0$ or $infty/infty$ only. Can this be done without L'Hospital in any way?

Answer 1

You had $|x|^{\sin(x)}$, whose limit is a $0^0$ form. You wrote it as $\exp \left ( \sin(x) \log(|x|) \right )$, which reduces the problem to finding the inner limit, which is a $0 \cdot \infty$ form. You can turn this into a $0/0$ form by writing $\sin(x) \log(|x|) = \frac{\sin(x)}{\frac{1}{\log(|x|)}}$. Similarly you can turn it into a $\infty/\infty$ form by writing $\sin(x) \log(|x|) = \frac{\log(|x|)}{\frac{1}{\sin(x)}}$. In either case you can now use L'Hospital's rule, and then apply $\exp$ to the result. I would suggest computing the left and right limits separately, so that the derivative of the $\log$ term makes sense.

Answer 2

It can be done if, and only if, you can prove that $$ \lim_{x \to 0} x \log x=0 $$ without using De l'Hospital. Can you?

By the way, it is possible, and many analysis books prove this by switching to exponentials, then to sequences, taking integer parts, and using some binomial expansion. I've always thought this approach is elementary but somehow hard to follow.