# Thread: How to prove differentiable at x is not 0

1. ## How to prove differentiable at x is not 0

Set
f(x) = {x sin(1/x), x not = 0
f(x) = {0, x = 0

and g(x) = xf(x)

Assume we know that f is continuous at 0 but not differentiable there, and that g is differentiable at 0. Both f and g are differentiable at each x not = 0.

(a) Find f'(x) and g'(x) for x not = 0.
(b) Show that g' is not continuous at 0.

I can find the f'(x) does not exist exist at 0 and g'(x) equals to 0, but I do not know how to prove something when says is not at a point, help me please.

2. Originally Posted by 450081592
Set
f(x) = {x sin(1/x), x not = 0
f(x) = {0, x = 0

and g(x) = xf(x)

Assume we know that f is continuous at 0 but not differentiable there, and that g is differentiable at 0. Both f and g are differentiable at each x not = 0.

(a) Find f'(x) and g'(x) for x not = 0.
(b) Show that g' is not continuous at 0.

I can find the f'(x) does not exist exist at 0 and g'(x) equals to 0, but I do not know how to prove something when says is not at a point, help me please.
At any point x apart from x=0 the functions are given by simple formulas, $f(x) = x\sin(1/x)$ and $g(x) = x^2\sin(1/x)$, and you can differentiate them by the usual rules (product rule, chain rule).

3. thanks now I know how to find f'(x) and g'(x) for x not equal to 0, but how do I prove g' is not continuous at 0, I really bad at proving something that is not something. I should prove by contradiction, right?

I have got f'(x) = sin(1/x) - 1/xcos(1/x) g'(x) = 2x sin(1/x)-cos(1/x), can I just show that the limit does not exist on g'(x) when x is approaching to 0?

4. Originally Posted by 450081592
thanks now I know how to find f'(x) and g'(x) for x not equal to 0, but how do I prove g' is not continuous at 0, I really bad at proving something that is not something. I should prove by contradiction, right?

I have got f'(x) = sin(1/x) - 1/xcos(1/x) g'(x) = 2x sin(1/x)-cos(1/x), can I just show that the limit does not exist on g'(x) when x is approaching to 0?
Exactly! The limit at 0 for g(x) doesn't exist because thus g'(x) can't exist where g(x) doesn't have a limit.

5. In particular, if a function is not continuous at x= a, it cannot be differentiable there.

6. Originally Posted by Jameson
Exactly! The limit at 0 for g(x) doesn't exist because thus g'(x) can't exist where g(x) doesn't have a limit.

There's a confusion here, me thinks: it's g'(x) which isn't continuous at x = 0. Of course g is cont. at x=0 since, as shown, it is differentiable there.
Now, as the OP wrote, $g'(x)=2x\sin \frac{1}{x}-\cos \frac{1}{x}$ , and it's easy to see the limit of this when x approaches zero doesn't exists. For example, one can make $x\rightarrow 0$ through the points $x_n=\frac{1}{2n\pi}\,,\,n\in \mathbb{N}$ and then through $y_n=\frac{1}{(2n-1)\pi}\,,\,n\in \mathbb{N}$ , and we get two different results.

Tonio

7. Originally Posted by tonio
There's a confusion here, me thinks: it's g'(x) which isn't continuous at x = 0. Of course g is cont. at x=0 since, as shown, it is differentiable there.
Now, as the OP wrote, $g'(x)=2x\sin \frac{1}{x}-\cos \frac{1}{x}$ , and it's easy to see the limit of this when x approaches zero doesn't exists. For example, one can make $x\rightarrow 0$ through the points $x_n=\frac{1}{2n\pi}\,,\,n\in \mathbb{N}$ and then through $y_n=\frac{1}{(2n-1)\pi}\,,\,n\in \mathbb{N}$ , and we get two different results.

Tonio
I didn't notice the existence of the limit xsin(1/x) or x^2sin(1/x), which both clearly exist. I wrote the limit doesn't exist because I was thinking of the difference quotient but wasn't thinking it through obviously. Thanks for pointing that out.