Does $\lim_{x \to a}\frac{f(x)}{g(x)}=0 \text{ and } \lim_{x \to a}g(x)=0 \implies \lim_{x \to a}f(x) = 0$? The Problem
Say
$$
\lim_{x \to a}\frac{f(x)}{g(x)}=0 \text{ and } \lim_{x \to a}g(x)=0 \, .
$$
Does this imply that $$\lim_{x \to a} f(x) = 0 \, ?$$
I understand that $f(x)$ must become very small relative to $g(x)$ as $x$ approaches $a$, but I'm worried that my intuitions might be failing me here.
What I have tried so far
For some neighbourhood of $a$, it seems evident that $|f(x)|\leq|g(x)|$. (Otherwise, the limit wouldn't equal $0$.) Consider the case where both $f(x)$ and $g(x)$ are nonnegative. Then, we are left with $0 \leq f(x) \leq g(x)$. Since
$$
\lim_{x \to a}g(x)=\lim_{x \to a}0=0 \, ,
$$
the squeeze theorem seems applicable here. However, I'm not sure how to generalise this result to when $|f(x)|\leq|g(x)|$, nor am I certain how to show that $|f(x)|\leq|g(x)|$ in the first place.
 A: You can rearrange $~f(x)~$as $~f(x)=\frac{f(x)}{g(x)}\cdot g(x)~$, then $$\lim_{x \to a} f(x) = \lim_{x \to a} \dfrac{f(x)}{g(x)}\cdot g(x) =\lim_{x \to a} \dfrac{f(x)}{g(x)}\cdot \lim_{x \to a}  g(x) = 0 \cdot 0=0~.$$
A: Overcomplicating things is not always bad. Maybe it will help you improve your $\epsilon-\delta$ proof skills. I'll prove the contrapositive by assuming $\neg \left(\lim_{x \to a} f(x) = 0\right)$ and $\lim_{x \to a}g(x)=0$, and show that $\neg \left(\lim_{x \to a}\frac{f(x)}{g(x)}=0\right)$. Let us examine what each of these statements means.
First, define $E_{z,\alpha} = (z-\alpha,z)\cup(z,z+\alpha).$
$\neg \left(\lim_{x \to a} f(x) = 0\right)$ is equivalent to the statement:
$\exists r>0$ such that $\forall \gamma>0,\ \exists \ x_\gamma\in  E_{a,\gamma}$  such that $|f(x_\gamma)-0| = |f(x_\gamma)| > r.$
Next, $\lim_{x \to a}g(x)=0$ is equivalent to the statement:
$\forall \varepsilon>0, \exists \delta>0$ such that $|f\left(x\right)-0| = |f\left(x\right)| <\varepsilon,\ \forall x \in E_{a,\delta}.$
We want to show that $\exists r'>0$ such that $\forall \beta > 0, \exists x_\beta \in E_{a,\beta}$ such that $\frac{|f(x_\beta)|}{|g(x_\beta)|} > r'.$
To this end, let $\beta > 0$ and then let
$
\beta'=
\begin{cases}
 \beta&\text{if}\, 0<\beta\leq 1\\
 \frac{1}{\beta}&\text{if}\, \beta > 1.\\
\end{cases}
$
Then $\exists \delta>0$ such that $ |g(x)| < \beta' \leq \beta \quad \forall \ x \in E_{a,\delta}$.
Also, $\exists x' \in E_{a,\delta}$ such that $|f(x')| > r.$
So now it makes sense to let $r' = r.$
Therefore, $\frac{|f(x')|}{|g(x')|} > \frac{r}{\beta'} \geq r$ for this $x' \in E_{a,\delta}.$
In other words, $r' = r$, and we have shown that $\exists r$ such that $\forall \beta > 0, \exists \delta > 0$ such that $\exists x' \in E_{a,\delta}$ with $\frac{|f(x')|}{|g(x')|} > r,$
i.e. $\lim_{x \to a} \frac{|f(x)|}{|g(x)|}$ does not exist.
