Question regarding derivatives

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March 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!
March 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.
March 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?
March 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.
March 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.