Prove differentiable implies continuous at x=xo

[splash_lover]

1. Prove f is differentiable at x=xo implies f is continuous at x=xo using epsilon and delta notation.

2. I have gotten this far:
absolute value(f(x)-f(xo)) <= absolute value(x-xo)*(epsilon + absolute value(f '(xo)))

<= means less than or equal to.


3. I need to get here:
absolute value(f(x)-f(xo)) <= epsilon

Could someone please help me? I think I have just looked at this problem for so long that I am missing the obvious. This problem is due tomorrow. So any help before then would be greatly apprieciated

 

[hunt_mat]

In terms of epsilon and delta the definition of a derivative is:

$$\left|\frac{f(x)-f(x_{0})}{x-x_{0}}-f'(x_{0})\right| <\varepsilon \quad |x-x_{0}|<\delta$$

Which means that:

$$(-\varepsilon +f'(x_{0})(x-x_{0})<f(x)-f(x_{0})<(\varepsilon +f'(x_{0})(x-x_{0})$$

Not it is a matter of redefining your epsilon.

 

[splash_lover]

but I need to get rid of the f'(x) and the (x-x0). How do I do that?

 

[hunt_mat]

if f'(x_0) is a fixed value, epsilon is small and then x-x_0 is small the f(x)-f(x_o) is between two very small numbers, so come up with a new epsilon