why is $\displaystyle \frac{\log(\sin x)}{\log(x)}$ $\quad\frac{\infty}{\infty}$ form as $x\to 0$?

Question

In this question $\displaystyle\frac{\log(\sin x)}{\log x}$ is taken as $\displaystyle\frac{\infty}{\infty}$ indeterminate form. But $\log(0)$ is not defined so how can L'Hospital's rule can be used?

Answer 1

Because the logarithm function approaches $-\infty$ as its argument approaches $0$ (from the right).

Answer 2

Remember that for small $x$ we know $x \approx \sin(x)$. This means that $$\lim_{x \to 0}\frac{\ln(\sin(x))}{\ln(x)} \approx \lim_{x \to 0} \frac{\ln(x)}{\ln(x)}$$ And since $\lim_{x \to 0}\ln(x) =- \infty$, it follows that $\lim_{x \to 0}\frac{\ln(\sin(x))}{\ln(x)}$ is of the form $\frac{-\infty}{-\infty}$.