4

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?

mathslover
  • 1,482
  • 5
    No, for the right hand side, the derivative could be negative. Consider $y=1/x^2$ around $x=0$ – imranfat Jun 22 '20 at 15:46
  • 1
    @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. – DonAntonio Jun 22 '20 at 15:51
  • @donantonio yes, that's a point. I addressed (or I think I did) what the OP wrote in bracketts (the infinite part) If the limit exists, then that's a separate case I didn't address. I wouldn't mind seeing someone else chime in – imranfat Jun 22 '20 at 15:59
  • 1
    @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. " – DonAntonio Jun 22 '20 at 16:18
  • 1
    It depends on what "in its natural domain" means. The imranfat example works if we define "the natural domain" to be $(0, 1)$. – Michael Jun 22 '20 at 16:21
  • Seems natural means $\mathbb {R}$ – aud098 Jun 22 '20 at 16:24
  • 1
    @DonAntonio Hello my friend. I hope that you are staying safe and healthy. My interpretation is literal: :"Is it always true that either $\lim_{x\to a} f'(x)>0$ if the limit exists or $\lim_{x\to a}f'(x)$ is infinite?". And we know the answer is "no" since it is possible that the limit fails to exist. – Mark Viola Jun 22 '20 at 16:26
  • Can you show us one example where the limit exists and equal to 0? – aud098 Jun 22 '20 at 16:35
  • @MarkViola Hellos and health. I 'm not sure that's accurate, but if the OP doesn't intervene we wopn't be sure. – DonAntonio Jun 22 '20 at 16:39
  • If $a$ is interior to the "natural domain" of the function, such as if $f:\mathbb{R}\rightarrow\mathbb{R}$ and $a \in \mathbb{R}$, then there is no way for $f$ to be continuous over its domain while still having $\lim_{x\rightarrow a} f(x) = \infty$. – Michael Jun 22 '20 at 16:41
  • 2
    @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. – DonAntonio Jun 22 '20 at 17:01
  • 1
    I think the two-sided limit of $f'$ cannot exist anyway. Did you mean a one-sided limit, $x\to a^-$? – mr_e_man Jun 22 '20 at 17:16
  • @mr_e_man Why so? – Vivaan Daga Jun 22 '20 at 18:40
  • @VivaanDaga - My answer shows that the left limit must be $+\infty$ if it exists; similarly the right limit must be $-\infty$ if it exists. – mr_e_man Jun 22 '20 at 18:44
  • Yes I meant $$ x \rightarrow a- .$$ My bad! The function f is continuous in it's domain expect for the point a, where it isn't defined. I suppose that a can be infinite as well. – mathslover Jun 22 '20 at 19:23
  • Found this duplicate: https://math.stackexchange.com/q/1863341/472818 – mr_e_man Jan 11 '23 at 21:07

2 Answers2

5

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

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

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

$$0<x-(a-\delta)=(x-a)+\delta<\delta,$$

so we have

$$f(x)-f(a-\delta)<B\cdot\big(x-a+\delta\big)<B\delta$$

$$f(x)<f(a-\delta)+B\delta.$$

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).

mr_e_man
  • 5,364
2

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: $$ \lim_{x \rightarrow \infty} \frac{1}{x} = 0 $$

Manuel Norman
  • 1,759
  • I don't think this is allowed as the original post said $;a;$ is a point where $;f;$ isn't defined – DonAntonio Jun 22 '20 at 16:15
  • 1
    I think the question is asking whether the limit of the derivative can be less than 0? – Rasputin Jun 22 '20 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".) – mr_e_man Jun 22 '20 at 19:40
  • @Rasputin the previous version of the question asked for the limit being $>0$ – Manuel Norman Jun 22 '20 at 19:40