I ran into a scenario when practicing L'Hôpital's rule which yielded -infinity/0. I broke this down into $-1 \cdot \infty \cdot \frac 1 0$, which I assumed equaled $-1\cdot\infty\cdot\infty$, which simplified to $-1\cdot\infty$ which equals negative infinity. So where did I go wrong with my logic as Wolfram Alpha claims the answer is positive infinity?
Here is the limit problem:
$$\lim_{x\to0}\frac{\ln(\sin x)}{\ln(\cos x)}$$
Wolfram Alpha does not say that $-\infty/-0 = \infty$ exactly, it says that this is equal to complex infinity. What's happening is that Wolfram Alpha is coming up with an interpretation for your inputs that makes the input sensible. Specifically, you can't divide infinity by zero in the context of real or complex numbers, but you can do this in the context of the Riemann sphere, which is usually treated as the union of the complex numbers with a single point at $\infty$.
As has already been pointed out in the comments above, $-\infty / 0$ is not a well defined expression. We are not allowed to divide by $0$ and we als have to remember that $\infty$ is not a number. That said, you can still find the limit. In asking that, we are just interested in what happens to the values of $f(x) = \frac{\ln(\sin(x))}{\ln(\cos(x))}$ when as $x$ approaches $0$.
First, I assume that $x$ is approaching $0$ from the right: $$ \lim_{x\to 0^+} \frac{\ln(\sin(x))}{\ln(\cos(x))} $$
since when $x$ is close to $0$ and negative, then $\sin(x)$ is negative and then $\ln(\sin(x))$ isn't defined.
So as $x$ approaches $0$ from the right, then $\sin(x)$ approaches $0$ from the right. Then $\ln(x)$ approaches $-\infty$.
Likewise, $\cos(x)$ approaches $1$ from the left, so $\ln(\cos(x))$ approaches $\ln(1) = 0$ from the left: $$ \ln(\cos(x)) \to 0^- \quad\text{as}\quad x\to 0^+ $$
And here then is the crucial part. As $x$ gets closer and closer to $x$ while being positive, we have just nooted that $\ln(\cos(x)$ is negative while $\ln(\sin(x))$ is negative. So for values of $x$ very close to $0$ but positive, $f(x)$ is something negative divided by something negative, hence positive.
Also, the bottom is close to $0$ (small number) while the top is large (and negative). That makes the whole thing very large and positive
Therefore The whole limit is positive $\infty$.
If you want a precise proof and you are willing to accept that $$ \ln(\sin(x)) \to -\infty \quad\text{as}\quad x\to 0^+ $$ and $$ \ln(\cos(x)) \to 0^- \quad\text{as}\quad x\to 0^+ $$ then let $N>0$ be given. We want to find a $\delta > 0$ such that if $0<x<\delta$ then $f(x) > N$. Now pick a $\delta_1$ such that $\ln(\sin(x)) < -N$ when ever $0<x<\delta_1$.
Pick a $\delta_2$ such that $0 > \ln(\cos(x)) > -1$ whenever $0<x<\delta_2$. Then for $\delta = \min\{\delta_1, \delta_2\}$ you will have $f(x) > N$ for $0<x< \delta$.
$$ \lim_{x\to 0^{+}}{\ln\left(\sin\left(x\right)\right) \over \ln\left(\cos\left(x\right)\right)} = \lim_{x\to 0^{+}}{\ln\left(x\right) \over \ln\left(1 - x^{2}/2\right)} = 2\lim_{x\to 0^{+}}{1 \over x^{2}}\,\left[-\ln\left(x\right)\right] = {\large +\,\infty} $$
has limit value $0$, but it gives an indeterminate form (notice Taylor series for $\cos x$), so we can't immediately know if the limit is equal to $0$. This is one method used in an answer.
– user263326 Sep 17 '17 at 18:23