is $f\left(x\right)\:=\:\left|x\right|^3$ twice differentiable? Consider the function $f\left(x\right)\:=\:\left|x\right|^3$ , $f:\mathbb{R}\rightarrow \mathbb{R}$.
1) Is it twice differentiable? And if so, how can I prove this and calculate it?
2) If it does, can I conclude it has third differentiable for every $x$?  
So far, for 1, I believe is true but I really stuck in the proof:
$$\lim _{h\to 0}\left(\frac{f\left(x+h\right)-f\left(x\right)}{h}\right)\:=\:\lim _{h\to 0}\left(\frac{\left|x+h\right|^3-\left|x\right|^3}{h}\right)\:=\:\lim_{h\to 0}\left(\frac{\left(\left|x+h\right|^{\:}-\left|x\right|\right)\left(\left|x+h\right|^2\cdot \left|x+h\right|\left|x\right|+\left|x\right|^2\right)}{h}\right)\:$$ But how to continue from here? I also tried this approach:  $$\lim _{x\to x_0}\left(\frac{f\left(x\right)-f\left(x_0\right)}{x-x_0}\right)\:=\:\lim_{x\to x_0}\left(\frac{\left|x\right|^3-\left|x_0\right|^3}{x-x_0}\right)\:=\:\lim_{x\to x_0}\left(\frac{\left(\left|x\right|^{\:}-\left|x_0\right|\right)\left(\left|x\right|^2+\left|x\right|\left|x_0\right|+\left|x_0\right|^2\right)}{x-x_0}\right)\:\:$$
I have serious problem with the next step. Can someone guide me what tricks I need to do? Thanks in advance!
 A: You can show that $f'(x) = 3x|x|$ by looking at the cases $x > 0$ (where $f(x) = x^3$), $x < 0$ (where $f(x) = -x^3$) and $x = 0$ (use the definition of the derivative). 
Then show that $f''(x) = 6|x|$ by looking at the cases $x > 0$ (where $f'(x) = 3x^2$), $x < 0$ (where $f'(x) = -3x^2$) and $x = 0$ (use the definition of the derivative again).  
Thus $f''(x)$ exists everywhere. Try this one more time and you will find that $f'''(0)$ does not exist. Hence $f'''$  does not exist everywhere. 
A: For $x\neq0$ the function is just $x^3$ or $-x^3$ in a neighborhood of the point. Being the derivative a local property the derivative is going to coincide with the derivative of $x^3$ or $-x^3$ according to the case. 
For $x=0$ you get to compute the limit $$\lim_{x\to0}\frac{f(x+0)-f(0)}{x-0}=\lim_{x\to0}\frac{|x|^3}{x}=\lim_{x\to0}\text{sgn}(x)|x^2|=0$$.
For the second derivative we need to look at one-sided limits $$\lim_{x\to0\pm}\frac{f'(x+0)-f'(0)}{x-0}=\lim_{x\to0\pm}\frac{f'(x)}{x}$$ Here $f'(x)$ is equal to $3x^2$ or $-3x^2$ according to the case. In both the limit is zero again.

A shorter way to get the derivatives at zero could be to notice that $$f(x)=x^2|x|=0+0x+0x^2+o(|x|^2).$$ But this requires knowing the relationship between Taylor with Lagrange remainder and the existence of derivatives.

A: Yes.
To see it graphically, here is the plot (generated here) of $f'(x) = 3 |x|^2 \frac{|x|}{x} \equiv 3 |x|^2 \operatorname{sign}(x)$:

As you can see, this function has well defined slope at $x=0$, and its derivative is zero at that point.
A: since the function $f$ is even, we only need to worry about the derivative for $x \ge 0.$ in this region $f(x) = x^3, f'(x) = 3x^2, f''(x) = 6x, f'''(x) = 6.$  checking the derivative at $x = 0,$ you see that they are all zero up to the second derivative, implying the function $f$ is twice differentiable for all $x.$ but,
the third derivative from the right is $3$ and from the left is $-3.$ this follows from the symmetry of $f.$ therefore $f$ is not three times differentiable at $x = 0.$ 
A: Does the function $f(x)=|x|^3$ has a Taylor expansion of order $2$ in $0$? 
You can write $f(x)=a+bx+cx^2+x^2\varepsilon(x)$ with $\varepsilon (x) \to 0$ as $x \to 0$. The answer is Yes, with $a,b,c=0,\ \varepsilon(x)= |x|^3/x^2$.
