1. ## continuous

Hi.

let

$f:X\rightarrow Y ~~and~~ there~~ exists~~ {f^'}_-(x_0),{f^'}_+(x_0)$

It is a continuous function f on x_0? is it true or false? please prove or example

thanks

2. ## Re: continuous

Is there some reason why you don't want to do this problem? Just start looking at some examples.

(I assume that $f'_-(x_0)$ means $\lim_{x\to x_0} f'(x)$ and equivalently for $f'_+(x_0)$.)

3. ## Re: continuous

Originally Posted by HallsofIvy
(I assume that $f'_-(x_0)$ means $\lim_{x\to x_0} f'(x)$ and equivalently for $f'_+(x_0)$.)
I would guess that $f'_-(x_0)=\lim_{x\to0^-}\frac{f(x+h)-f(x)}{h}$ (left limit). Maybe the OP can clarify.

4. ## Re: continuous

${f^'}_+(x_0)= \lim_{x\rightarrow {x_o}^+}...$

if both of them there exists, then f countious or not?

5. ## Re: continuous

This could mean $f'_+(x_0)=\lim_{x\to x_0^+}f'(x)$ or $f'_+(x_0)=\lim_{x\to x_0^+}\frac{f(x)-f(x_0)}{x-x_0}$, which are two very different things. Regardless, though, you should make some effort in answering the question.

6. ## Re: continuous

Originally Posted by emakarov
This could mean $f'_+(x_0)=\lim_{x\to x_0^+}f'(x)$ or $f'_+(x_0)=\lim_{x\to x_0^+}\frac{f(x)-f(x_0)}{x-x_0}$, which are two very different things. Regardless, though, you should make some effort in answering the question.
$f'_+(x_0)=\lim_{x\to x_0^+}\frac{f(x)-f(x_0)}{x-x_0}$

7. ## Re: continuous

I knew it!

It's a well-known fact that if f is differentiable at $x_0$, then f is continuous at $x_0$. Try to adapt the proof to this situation.

8. ## Re: continuous

Originally Posted by emakarov
I knew it!

It's a well-known fact that if f is differentiable at $x_0$, then f is continuous at $x_0$. Try to adapt the proof to this situation.
thanks, but it is say f^' + and f^' - there exist, maybe f^' not exist.. Is it possible?

9. ## Re: continuous

Yes, for example for |x| at x = 0.

Edit: But the proof of continuity is almost the same as in the case of the regular derivative.

10. ## Re: continuous

Originally Posted by emakarov
Yes, for example for |x| at x = 0.

Edit: But the proof of continuity is almost the same as in the case of the regular derivative.
thanks for your help, and Excuse me. I'm not good English

Well I did not answer my question
Now if the f^' + and f^' - there exists, then f will be countions? is it true?

i'm very sorry

11. ## Re: continuous

I realized that I myself got confused. No, if both one-sided derivatives exist, the function is not necessarily continuous. Both one-sided limits of the function exist (the proof of this is almost the same as for the fact that differentiable functions are continuous), but these limits are not necessarily the same. Consider an example shown here.