# Question regarding derivatives

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• Mar 10th 2010, 09:28 PM
Solid8Snake
Question regarding derivatives
Hello! I am having a lot of difficulty in terms of solving this problem, if anyone could help me that would be greatly appreciated. My teacher stated he would quiz us on this question, yet I have no clue how to approach it.

a) Set up the value of the function f(x)=xsin(1/x) at 0 to make it continuous. Then, by using the definition of the derivative, show that the new f is not differentiable at 0.

b) Determine whether the same is true for the function g(x)=x^2sin(1/x)-->after removing discontinuity at 0.

Thanks in Advance!
• Mar 10th 2010, 09:46 PM
Random Variable
Define f(x) to be zero when x=0.

Then $f'(0) = \lim_{x \to 0} \frac{f(x) - f(0)}{x-0} = \lim_{x \to 0} \frac{x \sin (1/x)}{x} = \lim_{x \to 0} \sin (1/x)$

Sin(1/x) oscillates widly as x approaches zero; therefore, f'(0) does not exist.

But g'(0) =0.
• Mar 10th 2010, 09:48 PM
Solid8Snake
for the last part, did you use integrals because we are not in that point of the course, is there any way else to prove it?
• Mar 10th 2010, 10:09 PM
Random Variable
Quote:

Originally Posted by Solid8Snake
for the last part, did you use integrals because we are not in that point of the course, is there any way else to prove it?

Proving that $\lim_{x \to 0} \sin (1/x)$ does not exist?

It involves finding a sequence ( $s_{n}$) that converges to some value c ( $s_{n} \ne c$), but $f(s_{n})$ does not converge. Such a proof is usually done in a real analysis class.
• Mar 11th 2010, 03:31 AM
HallsofIvy
Quote:

Originally Posted by Solid8Snake
for the last part, did you use integrals because we are not in that point of the course, is there any way else to prove it?

No, the integrals at the bottom are part of his "signature" and have nothing to do with the problem.