1

If $f(x)$ is a real function such that $f'(x) → 3$ as $x → 0$, does that mean $f'(0)$ exists? I ask this because of problem 9 on page 115 of Rudin's Principles of Mathematical Analysis. The actual question is slightly different:

Let $f$ be a continuous real function on $ℝ^1$, of which it is known that $f'(x)$ exists for all $x ≠ 0$ and that $f'(x) → 3$ as $x → 0$. Does it follow that $f'(0)$ exists?

I believe it does exist because the left and right limits are equal and the function itself is continuous and defined (is there more that must be met than this?). But if continuity is not known, then I think this conclusion is invalid. Would the following counterexample be valid if continuity is not assumed?

Let $f(x) = 3x - 5$, for $x < 0$ and $f(x) = 3x + 5$, for $x ≥ 0$. Then clearly $f'(x) → 3$ as $x → 0$, but $f'(0)$ does not exist because of the jump discontinuity on $f(x)$ at $x=0$.

  • 1
    http://math.stackexchange.com/questions/1462584/a-question-about-existance-of-derivative-of-function-at-zero/1462610#1462610 – Roberto Nunez Nov 14 '15 at 16:23

1 Answers1

0

Your counter example when the function is not continuous is right; however, you can directly prove the existence of $f'(0)$ when assuming the continuity by L'Hoptial's theorem for limits as follows

$$\mathop {\lim }\limits_{x \to 0} \frac{{f(x) - f(0)}}{{x - 0}} = \mathop {\lim }\limits_{x \to 0} \frac{{f'(x)}}{1} = \mathop {\lim }\limits_{x \to 0} f'(x) = 3$$

Hosein Rahnama
  • 14,882
  • But doesn't that assume $f(x) - f(0)$ is zero? But this should be fine because f(x) is a continuous function. Do I understand that correctly? –  Nov 14 '15 at 16:38
  • @zagadka314: In fact $\mathop {\lim }\limits_{x \to 0} \left( {f(x) - f(0)} \right) = 0$ due to continuity and $\mathop {\lim }\limits_{x \to 0} \left( {x - 0} \right) = 0$ and hence L'Hopital is applicable. Yes, I think you got it! :) – Hosein Rahnama Nov 14 '15 at 16:40
  • @zagadka314: Are you there? :) – Hosein Rahnama Nov 14 '15 at 16:51
  • I haven't downvoted anyone. I wait a few days before I vote on anything because I like to think about it before I vote or choose a best answer. I don't know why someone downvoted you. (I just upvoted you though anyway) –  Nov 14 '15 at 16:57
  • And someone deleted the other answer too, I guess –  Nov 14 '15 at 16:57
  • @zagadka314: OK, thanks. :) I am so sorry for these anonymous down-voters! :D – Hosein Rahnama Nov 14 '15 at 16:58
  • I honestly don't even like downvotes, on my profile I have given only one, ever. Thank you for your help. I wish I saw the comments on the other answer, maybe the downvote worried the person and he deleted it an left. –  Nov 14 '15 at 16:59
  • @zagadka314: I do believe you! :) Hope you be successful with your analysis course. :) – Hosein Rahnama Nov 14 '15 at 17:02
  • Avoiding L'Hopital, we can say the difference quotient above equals $f'(c_x) \to f'(0).$ – zhw. Nov 14 '15 at 19:44
  • @zhw: The problem is to prove that the limit exist! :) – Hosein Rahnama Nov 14 '15 at 20:39
  • I proved it. $(f(x) - f(0))/(x-0) = f'(c_x)$ by the MVT. As $x\to 0, c_x \to 0.$ Therefore $\lim f'(c_x) = 3.$ Done. – zhw. Nov 14 '15 at 20:45
  • @zhw: Sorry for that! :) I don't what MVT is! So I son't have any idea. :) – Hosein Rahnama Nov 14 '15 at 21:20
  • MVT = mean value theorem. – zhw. Nov 14 '15 at 21:20
  • @zhw: AHA! :D, Now, I got what you said! :) Yes, that would be a good proof too! :) – Hosein Rahnama Nov 14 '15 at 21:26