Suppose a real valued function $$h(x) = \frac{f(x)}{g(x)}$$. And we know that $$\lim_{x\to a} h(x) = L$$ Where $L$ is real and finite. We also know that $$\lim_{x\to a}g(x) = 0$$ Can we infer from this that $\lim_{x\to a}f(x) = 0$ ?
If $f(x)$ approaches a non zero number, then $h(x)$ approaches infinity. The only case where the limit does make sense is when it is a $\frac00$ indeterminate form.
Can there be any other cases? If not, then is my proof a good enough formal proof?
(The motivation for this question was from a problem which was of the form $\lim_{x\to a} \frac{f(x)}{g(x)} = L$, where $L$ was a finite number. It could be seen from the question that $\lim_{x\to a}g(x) = 0$. $f(x)$ on the other hand had certain constants in it, so the same could not be said. The objective was to find the value of that constant. I wondered if I could assume that the limit is in $\frac00$ form and apply L'Hopital's rule, as both $f(x)$ and $g(x)$ were differentiable.)
Best Answer
It is merely a consequence of the product rule of limits: \begin{align*} \lim f(x)=\lim\dfrac{f(x)}{g(x)}\cdot g(x)=\lim\dfrac{f(x)}{g(x)}\cdot\lim g(x)=0 \end{align*} provided that both later limits exist as real numbers.