# Derivative when function approaches infinity.

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?

• No, for the right hand side, the derivative could be negative. Consider $y=1/x^2$ around $x=0$ Jun 22, 2020 at 15:46
• @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. Jun 22, 2020 at 15:51
• @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. " Jun 22, 2020 at 16:18
• It depends on what "in its natural domain" means. The imranfat example works if we define "the natural domain" to be $(0, 1)$. Jun 22, 2020 at 16:21
• @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. Jun 22, 2020 at 17:01

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)

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 $$f'(x) for all $$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

$$\frac{f(x)-f(a-\delta)}{x-(a-\delta)}

Note that $$a-\delta implies

$$0

so we have

$$f(x)-f(a-\delta)

$$f(x)

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:

$$f(x)=\frac{1}{x^2}-5\sin\frac{1}{x^2}$$

(graph).

• Manuel Norman's answer shows that it's possible for the limit of the derivative to be 0 though, right? Jun 22, 2020 at 18:41
• I assumed that $a$ is finite. Jun 22, 2020 at 18:43
• Ahh ok, that makes sense Jun 22, 2020 at 18:49
• +1 for the well written answer and also the counter-example at the end. 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: $$\lim_{x \rightarrow \infty} \frac{1}{x} = 0$$

• I don't think this is allowed as the original post said $\;a\;$ is a point where $\;f\;$ isn't defined Jun 22, 2020 at 16:15
• I think the question is asking whether the limit of the derivative can be less than 0? Jun 22, 2020 at 18:39
• @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".) Jun 22, 2020 at 19:40
• @Rasputin the previous version of the question asked for the limit being $>0$ Jun 22, 2020 at 19:40