# Existence of a derivative

1. Aug 13, 2012

### sergey90

1. The problem statement, all variables and given/known data
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?

2. Relevant equations

3. 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.

2. Aug 13, 2012

### 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

3. Aug 13, 2012

### HallsofIvy

Staff Emeritus
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".

4. Aug 14, 2012

### Zondrina

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.