Let $f$ be a continous and differentiable function in it's natural domain.

If $$\lim_{x\to a-}f(x)=\infty$$, is it always true, that $$ \lim_{x\to a-} f'(x) \ge0 $$ if the limit exists (or is infinite)?

Intuitively this seems to be true, but can this be proven exactly?

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    $\begingroup$ No, for the right hand side, the derivative could be negative. Consider $y=1/x^2$ around $x=0$ $\endgroup$
    – imranfat
    Jun 22, 2020 at 15:46
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    $\begingroup$ @imranfat The problem with your example is that $\;\lim\limits_{x\to0}-\frac2x\;$ doesn't exist...and the OP is asking in case the limit of the derivative exists ...It is not a one-sided limit. $\endgroup$
    – DonAntonio
    Jun 22, 2020 at 15:51
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    $\begingroup$ @imranfat I think the OP meant $\;\lim f'\;$ always exists, even if the limit is infinite ,since some people believes "limit exists" necessarily mean "exists finitely ...anyway, I think the OP meant that the limit always exists. " $\endgroup$
    – DonAntonio
    Jun 22, 2020 at 16:18
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    $\begingroup$ It depends on what "in its natural domain" means. The imranfat example works if we define "the natural domain" to be $(0, 1)$. $\endgroup$
    – Michael
    Jun 22, 2020 at 16:21
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    $\begingroup$ @Michael Someone edited the post, but the original one said $\;a\;$ is not a point of definition of $\;f\;$ ...so in fact $\;x=a\;$ is a vertical asymptote. $\endgroup$
    – DonAntonio
    Jun 22, 2020 at 17:01

2 Answers 2


Define $L=\lim_{x\to a^-}f'(x)$, assuming this limit exists (or is infinite).

Suppose $L=-\infty$. Plug in any number $B$ to the definition of this limit, to get an interval $(a-\delta,a)$ within which $f'(x)$ is bounded above:

$$\exists B:\exists\delta>0:\forall x\in(a-\delta,a):f'(x)<B.$$

Suppose $L$ is finite. Plug in any number $\varepsilon>0$ to the definition of limit, and define $B=L+\varepsilon$, to get an interval $(a-\delta,a)$ within which $f'(x)$ is bounded:

$$\exists B>L:\exists\delta>0:\forall x\in(a-\delta,a):f'(x)\in(L-\varepsilon,L+\varepsilon)=(2L-B,B).$$

So, in any case with $L<+\infty$, we have $f'(x)<B$ for all $x$ close enough to $a$. (If $B$ is negative, then $f'(x)$ is also bounded by any positive number; so we can assume $B>0$.)

Now take any point $x$ within that interval. By the mean value theorem, $\frac{f(x)-f(a-\delta)}{x-(a-\delta)}$ is the derivative of $f$ at some point in that interval, so


Note that $a-\delta<x<a$ implies


so we have



This says that $f$ is bounded in that interval, which contradicts

$$\lim_{x\to a^-}f(x)=+\infty.$$

Therefore $L\not<+\infty$; either $L=\lim_{x\to a^-}f'(x)=+\infty$ or the limit doesn't exist.

Here's an example of a case where $f\to\infty$ but the limit of $f'$ doesn't exist:



  • $\begingroup$ Manuel Norman's answer shows that it's possible for the limit of the derivative to be 0 though, right? $\endgroup$
    – Rasputin
    Jun 22, 2020 at 18:41
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    $\begingroup$ I assumed that $a$ is finite. $\endgroup$
    – mr_e_man
    Jun 22, 2020 at 18:43
  • $\begingroup$ Ahh ok, that makes sense $\endgroup$
    – Rasputin
    Jun 22, 2020 at 18:49
  • $\begingroup$ +1 for the well written answer and also the counter-example at the end. $\endgroup$
    – Paramanand Singh
    Jun 23, 2020 at 7:44

Assume that $a$ is allowed to be infinity. Then, in general, the answer is no. For instance, let $f(x)= \ln x$. Then: $$ \lim_{x \rightarrow \infty} \ln x = \infty$$ but: $$ \lim_{x \rightarrow \infty} \frac{1}{x} = 0 $$

  • $\begingroup$ I don't think this is allowed as the original post said $\;a\;$ is a point where $\;f\;$ isn't defined $\endgroup$
    – DonAntonio
    Jun 22, 2020 at 16:15
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    $\begingroup$ I think the question is asking whether the limit of the derivative can be less than 0? $\endgroup$
    – Rasputin
    Jun 22, 2020 at 18:39
  • $\begingroup$ @Rasputin - If the limit is $\lim_{x\to\infty}f'(x)<0$ then the function is eventually constantly decreasing, which contradicts $\lim_{x\to\infty}f(x)=+\infty$. (I don't mean "constantly decreasing" as in "$f'(x)=c$", but rather "never increasing".) $\endgroup$
    – mr_e_man
    Jun 22, 2020 at 19:40
  • $\begingroup$ @Rasputin the previous version of the question asked for the limit being $>0$ $\endgroup$ Jun 22, 2020 at 19:40

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