Is the function F(x) = x^2*cos(pi/x) differentiable at x=0?

[Rishavutkarsh]

Homework Statement

F(x)=
x^2*{Cos(Pi/x)} when x!=0 (Not equal to)[/B]
0 when x=0
Is the function differentiable at x=0 ??

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

2. Homework Equations - Basic limit formulas.

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.

 

[HallsofIvy]

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.

 

[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 to2xcos(\pi/x)+sin(\pi/x) cannot be defined for x=0, but that's 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.

 

[pasmith]

Homework Statement

F(x)=
x^2*{Cos(Pi/x)} when x!=0 (Not equal to)[/B]
0 when x=0
Is the function differentiable at x=0 ??

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

2. Homework Equations - Basic limit formulas.

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.

The limit <br /> \lim_{x \to 0} f&#039;(x) does not exist. However the limit <br /> f&#039;(0) = \lim_{x \to 0} \frac{f(x)}{x} does exist. Thus f is differentiable at 0 but f&#039; is not continuous at 0.

 

[Rishavutkarsh]

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 to2xcos(\pi/x)+sin(\pi/x) cannot be defined for x=0, but that's 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.

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

The limit <br /> \lim_{x \to 0} f&#039;(x) does not exist. However the limit <br /> f&#039;(0) = \lim_{x \to 0} \frac{f(x)}{x} does exist. Thus f is differentiable at 0 but f&#039; is not continuous at 0.

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