Continuity and its definition

Question

Is it true that a function $f$ is continuous at point $a$ if the following holds? $$ \exists\varepsilon > 0 \ \ \ \exists\delta>0 \text{ such that }\ \ |x - a|<\delta \Rightarrow |\frac{f(x) - f(a)}{x - a}| < \varepsilon $$

I know that this is easily solved when we use the limit definition of derivatives, but can it be solved without using derivatives?

No. Your conclusion does not rule out $$\frac{f(x) - f(a)}{x - a}< -100$$ — GEdgar, Nov 10 '18 at 22:11
@MichaelHoppe I doubt that's a duplicate of the question you linked. This question is about proving that Lipschitz continuity implies continuity, which is not discussed in the question you linked. — Scientifica, Nov 17 '18 at 12:39
You must exclude $x=a$. That is, $0 < |x - a|<\delta \Rightarrow |\frac{f(x) - f(a)}{x - a}| < \varepsilon$. It means that $f$ is differentiable in $a$ with $f'(a) = 0$. — Paul Frost, Nov 17 '18 at 12:55

score 1 · Answer 1 · answered Nov 10 '18 at 19:37

1

Yes, it is true. In the younterval $(x-\delta,x+\delta)$, you have$$\bigl\lvert f(x)-f(a)\bigr\rvert\leqslant\varepsilon\lvert x-a\rvert.$$Therefore, $\lim_{x\to a} f(x)-f(a)=0$.

answered Nov 10 '18 at 19:37

José Carlos Santos

440,053

score 0 · Answer 2 · answered Nov 10 '18 at 19:36

0

Take $$f(x)=\sqrt{|x|}$$

it is continuous at $x=0$ but

$$\frac{f(x)-f(0)}{x-0}$$ is not bounded near $0$.

answered Nov 10 '18 at 19:36

hamam_Abdallah

63,719

1

The OP wrote “if”, not “only if”. – José Carlos Santos Nov 10 '18 at 19:37
@JoséCarlosSantos If it is differentiable, it is necessarily continuous. – hamam_Abdallah Nov 10 '18 at 20:07
Your comment in no way adresses the point that I raised. – José Carlos Santos Nov 10 '18 at 20:12

Continuity and its definition

2 Answers2