1. ## Derivative proof

Problem. Let $g(x)=x^{2}sin( \frac {1}{x} )$, find g'(0) and prove that g'(x) is not continuous at x=0.

My work:

$g'(0) = \lim _{x \rightarrow 0 } \frac {g(x)-g(0)}{x-0} = \lim _{x \rightarrow 0 } \frac {x^{2}sin( \frac {1}{x} )-g(0)}{x}$, but g(0) is undefined, so how do you find the derivative?

Thanks.

2. No, it is defined g(0)=0.

3. Well, then, I have $\lim _{x \rightarrow 0 } \frac {x^{2}sin ( \frac {1}{x} )}{x} = \lim _{x \rightarrow 0 } xsin( \frac {1}{x} )$

So g'(0) = 0?

4. Yes because $\left| x\sin \frac{1}{x} \right| \leq |x|$ now use squeeze theorem.

5. Now to prove g'(x) is not continuous at x = 0.

Pick a sequence $\{ x_{n} \}$ that converges to 0.

Consider $\lim _{n \rightarrow \infty } | g'( x_{n}) - 0 | = \lim _{n \rightarrow \infty } | \lim _{x \rightarrow x_{n}} \frac {g(x) - g(x_{n})}{x-x_{n}} |$

I think I'm stuck...

6. To show that $\sin \frac{1}{x}$ is not continous at zero (for example) a sequene say $x_n = \frac{1}{\pi n}$ which is zero. And pick another sequence like $x_n = \frac{1}{\frac{\pi}{2}+\pi n}$ which is 1. So it is not continous.