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.
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?
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, , and it's easy to see the limit of this when x approaches zero doesn't exists. For example, one can make through the points and then through , and we get two different results.
Tonio