# continuous

• Apr 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
• Apr 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 $\displaystyle f'_-(x_0)$ means $\displaystyle \lim_{x\to x_0} f'(x)$ and equivalently for $\displaystyle f'_+(x_0)$.)
• Apr 16th 2012, 05:45 AM
emakarov
Re: continuous
Quote:

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

I would guess that $\displaystyle f'_-(x_0)=\lim_{x\to0^-}\frac{f(x+h)-f(x)}{h}$ (left limit). Maybe the OP can clarify.
• Apr 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?
• Apr 16th 2012, 12:25 PM
emakarov
Re: continuous
This could mean $\displaystyle f'_+(x_0)=\lim_{x\to x_0^+}f'(x)$ or $\displaystyle f'_+(x_0)=\lim_{x\to x_0^+}\frac{f(x)-f(x_0)}{x-x_0}$, which are two very different things. Regardless, though, you should make some effort in answering the question.
• Apr 17th 2012, 10:13 AM
vernal
Re: continuous
Quote:

Originally Posted by emakarov
This could mean $\displaystyle f'_+(x_0)=\lim_{x\to x_0^+}f'(x)$ or $\displaystyle f'_+(x_0)=\lim_{x\to x_0^+}\frac{f(x)-f(x_0)}{x-x_0}$, 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}
• Apr 17th 2012, 10:42 AM
emakarov
Re: continuous
I knew it! (Smile)

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