Proving Differentiability of a Continuous Function at x=0

[DJBruce]

Homework Statement

"A real valued function, f, has the following property:

$$\left|f\right|$$ is differentiable at $$x=0$$​

Prove that if we specify that f is continuous at 0, then f is also differentiable at 0."

Homework Equations

Since $$\left|f \right|$$ is differentiable we know the following:
$$\lim_{x\to0} \frac{\left|f(x)\right|-\left|f(0)\right|}{x}$$ exists, which means:

$$\vee \epsilon>0 \exists \delta >0$$ st if $$\left|x\right|<\delta, \left|\frac{\left|f(x)\right|-\left|f(0)\right|}{x} - \left|f'(0)\right|\right|<\epsilon$$

Likewise, since f is continuous at 0 we know:

$$\vee \epsilon>0 \exists \delta >0$$ st if $$\left|x\right|<\delta, \left|f(x) - f(0)\right|<\epsilon$$

We want to show:

$$\lim_{x\to0} \frac{f(x)-f(0)}{x}$$ exists, which mean we want to show:

$$\vee \epsilon>0 \exists \delta >0$$ st if $$\left|x\right|<\delta, \left|\frac{f(x)-f(0)}{x} - f'(0)\right|<\epsilon$$

The Attempt at a Solution

I am not really sure where to go. I thought of using triangle inequality, but that would won't work since it implies the other direction.

 

[micromass]

Maybe you should split up in three cases:

1) f(0)>0
In this case, we have f=|f| on a neighbourhood of 0. So differentiability of f is guaranteed.

2) f(0)<0
Analogous.

3) f(0)=0
This is probably the hardest case. Try to show that the derivative of |f| in 0, must be 0. This will help...