Is $\lim_{x\to 0+} \frac{\ln(x)}{\ln(x)} = \frac{-\infty}{-\infty} = 1$

Question

$$\lim_{x\to 0^+} \sin(x)^\frac{1}{\ln(x)} = ... = \exp \left(\lim_{x\to 0^+} \frac{\ln(\frac{\sin x}{x}) + \ln(x)}{\ln(x)}\right)$$

Now, from continuity we can evaluate each term separately.

$\lim_{x\to 0^+} \ln(\frac{\sin x}{x}) = 0$

Therefore, we have:
$$\exp \left(\lim_{x\to 0^+} \frac{\ln(x)}{\ln(x)}\right)$$

Technically, we have in the exponent $\frac{t}{t} = 1$, where $t=\ln(x)$.
$e$ is indeed the answer, but are we allowed to do this direct arithmetic operation?
I mean, shouldn't it be solved using L'Hôpital's rule if we want to be rigorous?

I'm asking that because my instructor marked $\checkmark$ on this and didn't mention something is wrong about this.

So, I'm actually asking: When are we allowed to do this kind of operation?

Update:
Isn't it like claiming that $$\frac{-\infty}{-\infty} = 1$$

Thanks.

Answer 1

You can simplify ${\ln x} / {\ln x}$ to $1$ before taking the limit, and therefore there's no problem. But writing ${-\infty}/{-\infty}$ is nonsense.

Answer 2

The L'Hôpital's rule isn't more rigorous than others, nay, mathematicians generally prefer not to use this rule because it is seen as a mechanical way to solve a limit. Furthermore the rule, if it is used mechanically, not always allows to solve a limit. For example, with the limit: $$\lim_{x\rightarrow\infty}\frac{e^x}{\sqrt{x^2+1}}$$ this rule fails.