finding the limit of fraction with $\ln x$ and $\sin x$

Question

here's a limit that I'm struggling to calculate. $$\lim_{x\to0} \frac{x^2\ln x}{\sin x(e^x-1)}$$ here's my work: $$\lim_{x\to0} \frac{x^2\ln x}{\sin x(e^x-1)} \iff \lim_{x\to0} \ln x/(\sin x(e^x-1))/x^2$$ I'm not sure on whether I should separate the $\sin x$ or not.

Answer 1

For $x>0$,$$\frac{x^2\ln(x)}{\sin(x)(e^x-1)}=\underbrace{\frac{x}{\sin(x)}}_{\to 1}\cdot \underbrace{\frac{x}{e^x-1}}_{\to 1}\cdot \underbrace{\ln(x)}_{\to -\infty }.$$