# Definition of Continuity?

1. Apr 11, 2010

### sponsoredwalk

Hi I have a question about continuity as it is proven by the existence of a derivative.

The proof I've read is the following and I really just want to talk about it to be 100% sure I've understood it and I know where it comes from;

1: We'll take the equation of a line; $$f(x) \ - \ f(x_0) \ = \ m (x \ - \ x_0) \$$

where;

$$m \ = \ \frac{f(x) \ - \ f(x_0)}{x \ - \ x_0}$$

and set it up like so;

2: $$f(x) \ = \ \frac{f(x) \ - \ f(x_0)}{x \ - \ x_0} (x \ - \ x_0 ) \ + \ f(x)$$

So, as I understand the proof, if we want to prove that the function f is continuous we will take the limit of the above expression and as x tends back onto itself and by getting the equality f(x_0) = f(x_0) we've proved the function continues along at this point

3: $$\lim_{x \to x_0} f(x) \ = \ \lim_{x \to x_0} \frac{f(x) \ - \ f(x_0)}{x \ - \ x_0} (x \ - \ x_0 ) \ + \ \lim_{x \to x_0} f(x)$$

4: $$f(x_0) \ = \ f ' (x_0) (x_0 \ - \ x_0 ) \ + \ f(x_0)$$

5: $$f(x_0) \ = \ f ' (x_0) (0) \ + \ f(x_0)$$

6: $$f(x_0) \ = \ f(x_0)$$

I wonder how deep a proof this is, it is from an advanced calculus text. :tongue2:

2. Apr 11, 2010

### LCKurtz

I think that argument is very confused as it is written. Try looking at it this way (I'm going to use a instead of x0 to avoid the subscripts). Begin with this identity:

$$\frac{f(x)-f(a)}{x-a}= f'(a)+\frac{f(x)-f(a)}{x-a}-f'(a)$$

That is obviously an identity. Now multiply both sides by (x-a) ane rewrite it as:

$$f(x)-f(a)= f'(a)(x-a)+\left( \ \frac{f(x)-f(a)}{x-a}-f'(a)\ \right)(x-a)$$

Now take the limit as $x \rightarrow a$ on both sides. On the right side the quantity in the large parentheses goes to zero because the function is differentiable at a and the difference quotient approaches f'(a). The other two terms on the right obviously go to zero too. Therefore the left side goes to zero too and f is continuous at a.

3. Apr 11, 2010

### sponsoredwalk

But I don't understand, what you've essentially written the before and after product of a derivative in the first latex frame and then wrote something that just equals zero in the second latex frame.

The L.H.S. of your second latex frame is zero because as x-->a it becomes f(a) - f(a) ergo zero. The R.H.S. obviously goes to zero too because of (x - a) becoming (a - a) = 0. It just comes out of nowhere to me and could be speaking about anything.

As I understand the proof I've put up it defines the function to be continuous by employing the equation of a line and then attaining equality on both sides. It also works because it is applicable to any function directly.

I'm just very familiar with the form I've put up as the mean value theorem and newtons method jump directly from it.

Can you tell me what I'm confused about in writing it this way?

4. Apr 11, 2010

### LCKurtz

No. You are trying to prove that f is continuous at x = a. You aren't given that $f(x) \rightarrow f(a)$ so you can't assume it and you don't have the left side goes to zero given. The point of this argument is that you can prove $f(x) \rightarrow f(a)$ by noting that f(x) - f(a) is equal to an expression on the right side which obviously does go to zero. That is how you know $f(x) \rightarrow f(a)$ and that is how you prove f is continuous at a.

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