# Thread: prove that f is differentiable at x_0 and that f' is continuous at x_0

1. ## prove that f is differentiable at x_0 and that f' is continuous at x_0

suppose that f is continuous on (a,b). If f is also differentiable on (a,b), except possibly at a single point $\displaystyle x_0$ where lim_x->$\displaystyle x_0$ f'(x) exists and is finite, prove that f is differentiable at $\displaystyle x_0$ and that f' is continuous at $\displaystyle x_0$.

2. Originally Posted by tn11631
suppose that f is continuous on (a,b). If f is also differentiable on (a,b), except possibly at a single point $\displaystyle x_0$ where lim_x->$\displaystyle x_0$ f'(x) exists and is finite, prove that f is differentiable at $\displaystyle x_0$ and that f' is continuous at $\displaystyle x_0$.
$\displaystyle f'(x_0)=\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}=\lim_{x\to x_0}f'(x)$. Why?

3. Originally Posted by Drexel28
$\displaystyle f'(x_0)=\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}=\lim_{x\to x_0}f'(x)$. Why?
not really. basically we are asked to prove $\displaystyle \lim_{\delta\to 0}\lim_{h\to0}\frac{f(x_0+\delta+h)-f(x_0+\delta)}{h}=\lim_{h\to0} \lim_{\delta\to 0} \frac{f(x_0+\delta+h)-f(x_0+\delta)}{h}$, and the conclusions will follow. to show that you can interchange the limits, probably some standard results should apply (it feels that uniform boundedness of $\displaystyle f'$ is precisely what is needed)

4. Originally Posted by choovuck
not really. basically we are asked to prove $\displaystyle \lim_{\delta\to 0}\lim_{h\to0}\frac{f(x_0+\delta+h)-f(x_0+\delta)}{h}=\lim_{h\to0} \lim_{\delta\to 0} \frac{f(x_0+\delta+h)-f(x_0+\delta)}{h}$, and the conclusions will follow. to show that you can interchange the limits, probably some standard results should apply (it feels that uniform boundedness of $\displaystyle f'$ is precisely what is needed)
Oh, is that so? What makes you think what I said was incorrect?

5. Originally Posted by Drexel28
$\displaystyle f'(x_0)=\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}=\lim_{x\to x_0}f'(x)$. Why?
b/c of the mean value theorem?

6. Originally Posted by tn11631
b/c of the mean value theorem?
Haha, well. Very indirectly. Think more "limity"

7. Originally Posted by Drexel28
Haha, well. Very indirectly. Think more "limity"
hmm..I'm drawing a blank

8. Originally Posted by tn11631
hmm..I'm drawing a blank
What is $\displaystyle \lim_{x\to 0}\frac{\sin(x)}{x}$? How do you know? I bet it isn't because you expanded it with a Maclaurin series or used a geometrically based squeeze theorem argument.

9. Originally Posted by Drexel28
What is $\displaystyle \lim_{x\to 0}\frac{\sin(x)}{x}$? How do you know? I bet it isn't because you expanded it with a Maclaurin series or used a geometrically based squeeze theorem argument.
1? and im not sure if your using sarcasm lol but what about l'hopital's rule?

10. Originally Posted by tn11631
Mhmm!

11. Originally Posted by Drexel28
Mhmm!
hmm, i see that with the sinx/x but I'm confused at how to use it with the orig. problem lol

12. Originally Posted by tn11631
hmm, i see that with the sinx/x but I'm confused at how to use it with the orig. problem lol
Look at L'hopital's rule. $\displaystyle f(x)-f(x_0)\to 0$ (by continuity), and $\displaystyle x-x_0\to 0$ (obviously) and so...

13. Originally Posted by Drexel28
Look at L'hopital's rule. $\displaystyle f(x)-f(x_0)\to 0$ (by continuity), and $\displaystyle x-x_0\to 0$ (obviously) and so...
after rereading my question over and over again, I feel like just proving L'hopital's rule is what I have to do. but what about g?

14. Originally Posted by tn11631
after rereading my question over and over again, I feel like just proving L'hopital's rule is what I have to do. but what about g?
Try rereading it. The derivative exists everywhere except for maybe at that point, so L'hopital's applies and when you differentiate the numerator and denominator you get $\displaystyle \lim_{x\to x_0}f'(x)$...which...by...assumption...exists.

15. Originally Posted by Drexel28
Try rereading it. The derivative exists everywhere except for maybe at that point, so L'hopital's applies and when you differentiate the numerator and denominator you get $\displaystyle \lim_{x\to x_0}f'(x)$...which...by...assumption...exists.
right, i understand that. But wait isn't it telling us that the limit as x->x_0 f'(x) does exist and is finite? Aye, im confusing myself, i think.

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