differentiable/continuous?

Question

i couldnt solve this. just curious if anyone else could... :)


f(x) = x^2 (sin(1/x)) when x does not = 0
0 when x=0

a) using the definition of the derivative, prove that f is differentiable at x=0

b)show that f ' (x) is not continuous at x =0.

Answer 1

For x <> 0 we have (f(x) - f(0))/(x-0) = (x^2 sin(1/x))/x = x sin(1/x). The sine function is bounded, because its values are in [-1, 1]. Therefore, |x sin(1/x)| = |x| |sin(1/x)| <= |x| for every x<>0. It follows that lim (x -> 0) x sin(1/x) = 0. Therefore, by the definition of derivative, we have f`(0) = 0 , proving f is differentiable at x = 0.

For x<>0, the differentiation rules imply that f`(x) = x^2 cos(1/x) (-1/x^2) + 2x sin(1/x) = - cos(1/x) + 2x sin(1/x). cos(1/x) oscillates between -1 and 1 and has no limit as x-> 0; and 2x sin(1/x) -> 0 as x -> 0. Therefore their sum f`(x) = - cos(1/x) + 2x sin(1/x). has no limit as x-> 0. Since lim (x -> 0) f`(x) = f`(0) is a necessary condition for continuity at o, it follows f` is not continuous at x =0.,

Answer 2

Have you tried using approximation?
By approximation you could just use 2 points where x is so small that it could be close enough to 0 (i.e. +/- 0.000001 and 0.000002) . Then just get the tangent of those two points. Make sure you do it first with 2 points in the possitive side to see what is the tangent (which is actually the derivative) as it approaches 0 from the right. Then do the same with two points in the negative side to see what is the derivative as it approaches 0 from the left.

It won't be continuous since the point where x=0 doesn't exist (even though we can guess were it is), so your curve is broken in that point. You can't divide by zero.

Answer 3

f'(x) = 2x*sin(1/x)-cos(1/x)

a) For f to be differentiable at x=0, the derivative must exist and be defined at x=0. Since 1/x is undefined for x=0, f is not differentiable at x=0.

b) For f' to be continuous at x=0, the L.H limit must equal the R.H limit and f'(0) must be defined. Since f' is undefined at x=0, it is not continuous even though the LH limit = RH limit.

Pay no attention to gimb1960 - he does not know what he is talking about.

Answer 4

a)
well the left hand limit and right hand limit for x-> 0 of f(0) - f(x)/x is 0

b) huh ? not continous ? it is continous !