Existence of Derivative at x=0

[sergey90]

Homework Statement

f(x)= xsin(1/x) if x!=0
= 0 if x=0

does the derivative exist at x=0?

Can somebody please provide a visual backup of the result? Is this supposed to be a cusp that's why there is no derivative on a continuous function?

Homework Equations

The Attempt at a Solution

Using the squeeze theorem we see that the function is continuous at 0, but when we compute the derivative limit, we are left with limit[h->0]sin(1/h) which doesn't exist.

 

[Bacle2]

Right, the limit does not exist, so f is not differentiable at 0.

Why don't you try an online graphing calculator, e.g:http://www.meta-calculator.com/online/

You can zoom-in for better info.

Note that (h^2)sin(1/h) ; 0 at 0 is differentiable everywhere--limit at 0 is

h*sin(1/h)-->0

 

[HallsofIvy]

The derivative at 0 is given by
$$\lim_{h\to 0}\frac{f(h)- f(0)}{h}=\lim_{h\to 0}\frac{h sin(1/h)}{h}= \lim_{h\to 0}sin(1/h)$$
As h goes to 0, 1/h goes to infinity so sin(1/h) alternates and does not have a limit.

No, there is no "cusp". That occurs when you have differing limits from the right and left. Here, there is no "limit from the left" or "limit from the right".

 

[STEMucator]

Recall that : |sinx| ≤ 1 $\forall$x$\in$$\Re$

Which means : -1 ≤ sinx ≤ 1 $\forall$x$\in$$\Re$

Lets say f(x) = sinx, you now have f(x) bounded between h(x) = -1 and g(x) = 1. Now think about the squeeze theorem.