Will the L'Hopital's Rule work here, when it agrees with the derivation of the rule as I know it?

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Say I have to find the limit for:

$$\lim_{x\rightarrow a}\frac{f(x)}{g(x)}$$

Such that $$g(a)=0≠f(a)$$

Multiplying the numerator and denominator by $g(x)$, I get$$\lim_{x\rightarrow a}\frac{f(x)g(x)}{g(x)^2}$$ and I get the indeterminate form $\frac 00$. So can I use the L'Hopital's Rule to find the limit here?

From 3B1B's video about L'Hopital's Rule at time stamp 14:30, the rule works as the value of the numerator graph and denominator graph at given value which $x$ approaches is $0$, so I can just divide their derivatives. So will the L'Hopital's Rule work here, considering I can just derive the rule from such arrangement of the graphs?

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Say I have to find the limit for: $$\lim_{x\rightarrow a}\frac{f(x)}{g(x)}$$ Such that $$g(a)=0≠f(a)$$

If $f(x)$ and $g(x)$ are both continuous at $x=a$, then the situation has to be game over. That is, the limit must be $\pm \infty$.

This means that it does not matter what creative steps you attempt, with or without L'Hopital's rule. Creativity won't alter that the limit must be $\pm \infty$.

The reason is that under the assumption that both functions are continuous, you have $g(x)$ approaching $0$ and $f(x)$ approaching some non-zero number, as $x$ approaches $a$.


Thanks to Greg Martin and Arturo Magidin for indicating an oversight in my answer. As they have indicated, I overlooked that rather than the limit being $\pm \infty$ another possibility is that the limit might not exist.

See the comments following this answer, including the comment of Torsten Schoeneberg, which discusses the possibility of $f$ and/or $g$ not being continuous at $x=a$.