Prove differentiable implies continuous at x=xo

  • #1
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
 

Answers and Replies

  • #2
hunt_mat
Homework Helper
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In terms of epsilon and delta the definition of a derivative is:

[tex]
\left|\frac{f(x)-f(x_{0})}{x-x_{0}}-f'(x_{0})\right| <\varepsilon \quad |x-x_{0}|<\delta
[/tex]

Which means that:

[tex]
(-\varepsilon +f'(x_{0})(x-x_{0})<f(x)-f(x_{0})<(\varepsilon +f'(x_{0})(x-x_{0})
[/tex]

Not it is a matter of redefining your epsilon.
 
  • #3
but I need to get rid of the f'(x) and the (x-x0). How do I do that?
 
  • #4
hunt_mat
Homework Helper
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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
 

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