# Differentiable implies continuous

1. Aug 19, 2010

ehhh

Last edited: Aug 19, 2010
2. Aug 19, 2010

### Staff: Mentor

Is this a homework problem?

This is really hard to read, with everything on one line, and the malformed fraction near the beginning. To make it more readable, use a pair of tex and /tex tags for each line.

3. Aug 19, 2010

It wasn't responding to any of the changes I was making to the Latex code and I gave up. I'm not sure what was wrong. And it wasn't a homework problem. I've never taken analysis. I was just wondering if this proof works after reading the theorem.

4. Aug 19, 2010

### HallsofIvy

Staff Emeritus
Often you will find that you need to click on your web-reader's "refresh" button when you edit LaTex. Why it doesn't refresh when you save the edit, I don't know.

I cannot, of course, see what you have deleted so I cannot directly answer your question (which apparently is about a specific proof that "differentiable implies continuous") but I can say this:
f(x) is differentiable at x= a if and only if
$$\lim_{h\to 0}\frac{f(a+ h)- f(a)}{h}$$
exists. Since the denominator necessarily goes to 0, in order that the limit exist, the numerator must also go to 0: we must have
$$\lim_{h\to 0} f(a+h)- f(a)= 0$$
If we let a+h= x then as h goes to 0, x goes to a so that is the same as
$$\lim_{x\to a}f(x)- f(a)= 0$$

which is the same as
$$\lim_{x\to a} f(x)= f(a)$$
For that to be true, we must have
1) $\lim_{x\to a} f(x)$ exist
2) f(a) exist
3) the two are equal
which is precisely the definition of "continuous at x= a".

5. Aug 19, 2010

$$f'(c) = \lim_{x\to c}\frac{f(x) - f(c)}{x - c}$$ ...since it is differentiable at any arbitrary point.

$$f'(c) = \frac{\lim_{x\to c}f(x) - \lim_{x\to c}f(c)}{\lim_{x\to c}(x - c)}$$ ...using properties of the limit (i think).

$$f'(c) * \lim_{x\to c}(x - c) = f'(c) * 0 = 0 = \lim_{x\to c}f(x) - \lim_{x\to c}f(c)$$ ...the limit of x - c is 0. The limit of f(c) is f(c) so this implies that

$$\lim_{x\to c}f(x) = f(c)$$ which proves continuity.

I know this can't be right. I am just trying to learn some of this. Also, would a more rigorous proof than the one you posted be at all better? I've seen some very confusing ones with deltas and epsilons and wonder if this is overkill?

Last edited: Aug 19, 2010
6. Aug 19, 2010

### slider142

This does not work, as the limit of the denominator is 0. Distributing the limit over a quotient is only valid if the limit exists and the limit of the denominator is non-zero. However, you can get a working proof if you multiply both sides by $\lim_{x\rightarrow c} x - c$.
Going down to the level of deltas and epsilons is not necessary. There is a proof using only what you know about the definition of the derivative at a point, and the definition of continuous at a point, without explicitly using epsilon-delta form.

Last edited: Aug 19, 2010
7. Aug 20, 2010

OK, I think I see that. After that you can pull out the limit sign and cancel out the x - c and are left with $$0 = \lim_{x\to c}f(x) - f(c)$$. The proof is basically done.