continuous

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April 16th 2012, 02:00 AM
vernal

continuous

Hi.

let

http://latex.codecogs.com/gif.latex?...%7D_+%28x_0%29

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

thanks
April 16th 2012, 05:40 AM
HallsofIvy

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 $f'_+(x_0)$ .)
April 16th 2012, 05:45 AM
emakarov

Re: continuous

Quote:

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.
April 16th 2012, 09:24 AM
vernal

Re: continuous

http://latex.codecogs.com/gif.latex?...%20{x_o}^+}...

if both of them there exists, then f countious or not?
April 16th 2012, 12:25 PM
emakarov

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.
April 17th 2012, 10:13 AM
vernal

Re: continuous

Quote:

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.

http://latex.codecogs.com/png.latex?...x_0%29}{x-x_0}
April 17th 2012, 10:42 AM
emakarov

Re: continuous

I knew it! (Smile)

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.
April 17th 2012, 11:38 AM
vernal

Re: continuous

Quote:

Originally Posted by emakarov

I knew it! (Smile)

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?
April 17th 2012, 11:43 AM
emakarov

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.
April 17th 2012, 11:54 AM
vernal

Re: continuous

Quote:

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
April 17th 2012, 12:03 PM
emakarov

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.