differentiability at point

consider the function: if x . 0 if x=0. show f is differentiable at 0? is f continuous at 0? if f'(x) continuous at 0?. ... ok that is the problem. but i dont know how to take the limit since it is cos(1/h) and then you cant just stick in the zero. i cant use l'hopital which is what i usually try. using the limit of differentiability i get the derivative is hcos(1/h) as h goes to 0. but i cant do anything with it.

Re: differentiability at point

Quote:

Originally Posted by

**learning** using the limit of differentiability i get the derivative is hcos(1/h) as h goes to 0. but i cant do anything with it.

The limit of differentiability, eh? So, differentiability is a function of which you can take a limit, right? Do you know the value of differentiability at 5?

Concerning the limit of as , what happens to a magnitude that is locked between -1 and 1 as you multiply it by smaller and smaller values?

Re: differentiability at point

come on now you ok i did mistype. sorry. i mean limit definition of differentiability (at a point/general. at a point here). as for your remark on this limit.. i know the limit is zero and i know why. but how can i state it to be so rather just by intuition? i get no mark if i just say it is obvious. ok so since the limit of h as h goes to zero, you are multiplying a 0 by .... well anything would make it zero. BUT, the limit of cos(1/h) as h goes to zero doesnt exist so i cant just use product of limits = limit of products.

Re: differentiability at point

Re: differentiability at point

If you have an intuitive understanding why tends to 0 (as a product of a function that tends to 0 and a bounded function), it is easy to write an epsilon-delta proof that the limit is zero. Let's proof a generalization.

Theorem. Let f(x) be such that as and let g(x) be bounded in some neighborhood of . The latter means that there exist positive constants r and M such that implies for all x. Then as .

Proof. By definition of limit, for every there exist a such that implies for all x. Suppose some is given. Fix the corresponding . What happens to when both and , i.e., when ? Then . We would like to make to conform to the definition of limit. For this, we choose the that corresponds not to but to . Then implies and implies for all x.

Altogether, given an , there exists a , namely, where is obtained from the definition of limit of f(x) for . For this it is true that implies for all x.

You are left with an exercise to understand this proof, to apply it to the current problem and simplify where possible for concrete and .