# A question on limits?

1. Apr 16, 2015

### Rishavutkarsh

1. The problem statement, all variables and given/known data
F(x)=
x^2*{Cos(Pi/x)} when x!=0 (Not equal to)

0 when x=0
Is the function differentiable at x=0 ??

P.S.- Cos(pi/x) is under mod (Absolute value)

2. Relevant equations - Basic limit formulas.

3. The attempt at a solution
I tried to differentiate it wrt to x but I am getting the function to be non differentiable
However with first principle differentiation I get it as differentiable. Where am I going wrong? Any help appreciated.

2. Apr 16, 2015

### HallsofIvy

Staff Emeritus
I am confused as to what you want to show. To show "f is differentiable at x= 0" you want to show that $\lim_{h\to 0} \frac{f(h)- f(0)}{h}$ exits. That is not necessarily dependent on the derivative of f for x not 0.

3. Apr 16, 2015

### Delta²

you should just try to find the derivative at x=0 by applying the definition of derivative as a limit of a ratio. So you basically looking if the limit $\lim\limits_{x \to 0}xcos(\pi/x)$ exists.

I think what confuses you is that the derivative which in the interval $(-\infty,0)+(0,\infty)$ is equal to$2xcos(\pi/x)+sin(\pi/x)$ cannot be defined for x=0, but thats the point we ve excluded 0 in order to be able to do these calculations for the derivative wrt x. For x=0 you just have to follow the definition of derivative as a limit of the ratio.

Last edited: Apr 16, 2015
4. Apr 16, 2015

### pasmith

The limit $$\lim_{x \to 0} f'(x)$$ does not exist. However the limit $$f'(0) = \lim_{x \to 0} \frac{f(x)}{x}$$ does exist. Thus $f$ is differentiable at 0 but $f'$ is not continuous at 0.

5. Apr 17, 2015

### Rishavutkarsh

That was the exact point of confusion, I think I got it now though thanks.

This was the exact thing I needed to know, thanks.