# Function such that f'(x) exists but limf'(x) as x -> 0 does not

1. Apr 5, 2013

### physman123

1. The problem statement, all variables and given/known data
Does anyone have an example of a function such that f'(x) exists everywhere, but limf'(x) as x approaches 0 does not?

2. Relevant equations

3. The attempt at a solution
I was thinking something along the lines of f(x) = (2x^5 + 5x^2)/x^11
Since then f'(x) = (10x^4 + 10x) / 11x^10
and limf'(x) as x approaches 0 does not exist since 11(0)^10 = 0 and you cannot divide by 0. However, I realized that that means that f'(x) does not exist everywhere since limf'(x) doesn't exist as x approaches 0, that implies that f'(0) does not exist.

2. Apr 5, 2013

### Staff: Mentor

I see two problems with this.
1. Your function f is not defined at x = 0, so f' is also undefined at x = 0. This violates the condition that f' is defined everywhere.
2. Your differentiation is incorrect. The derivative of a quotient is NOT the quotient of the derivatives. To differentiate a quotient you normally need to use the quotient rule. For your example, carrying out the division first before differentiating makes it unnecessary to use the quotient rule.

f(x) = (2x5 + 5x2)/x11 = 2x-6 + 5x-9. Now you can differentiate using the sum rule and the power rule.

In any case, this is not a good example, since f' is not defined everywhere.

3. Apr 5, 2013

### Zondrina

So you want a function that's continuous everywhere, but not differentiable at 0.

How about f(x) = |x|

4. Apr 5, 2013

### Staff: Mentor

This one came to my mind, as well, but it doesn't work. For this function, f' doesn't exist at x = 0. The condition in the problem is that f'(x) exists everywhere.

5. Apr 5, 2013

### jbunniii

Hint: Try starting with $\sin(1/x)$ and multiplying by something suitable so that $f'(0)$ exists but $\lim_{x \rightarrow 0} f'(x)$ does not exist.

6. Apr 5, 2013

### Bacle2

f(x)=-x2/2 , for x<0

f(0)=0

f(x)=x2/2 for x>0

Then f'(x)=|x| , so that f' is defined everywhere, but the limits at 0 don't agree.

7. Apr 5, 2013

### jbunniii

But this $f'$ is continuous everywhere, so the limits will agree everywhere.