So I was wondering. Let $f:\mathbb{R} \supseteq U \rightarrow \mathbb{R} $ be a function and $p \in \mathbb{R} \cup \left\{ - \infty, +\infty \right\}$ such that

$$\lim_{x \rightarrow p} f(x) = 0$$

Does it now hold that $$\forall a \in \mathbb{R}\setminus \left\{ 0\right\}: \lim_{x \rightarrow p} \frac{a}{f(x)} = \pm \infty$$

That is, the limit $\lim_{x \rightarrow p} \frac{a}{f(x)}$ goes either to infinity or to minus infinity.

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    $\begingroup$ Even if $a\ne0$, we can have both $\pm\infty$ as limit points of $\frac a{f(x)}$, think e.g. $f(x) =1/x$. $\endgroup$ – Berci Jun 3 '18 at 9:21
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    $\begingroup$ Or maybe better $p=+\infty$, $U=(0,\infty)$, $f(x)=\frac1x\sin x$. Or simply $U=\Bbb R$, $f(x)=x$ and $p=0$. $\endgroup$ – Hagen von Eitzen Jun 3 '18 at 9:25
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    $\begingroup$ I think you need to add the condition that $f(x)>0$(or $<0$) for some deleted neighbourhood of $p$, otherwise there might not be a single limit point $\endgroup$ – Holo Jun 3 '18 at 9:35

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