Compute the gradient at the origin and show that it is continuous at the origin Let $f(x) = |x|^{p}, x \in \mathbb{R}^n$ with $p\geq2$. I have no idea of how to compute the gradient of this function at the origin. The only idea that I had is the computation of the limit of the partial derivatives , but I don't see how to compute the limit. Someone could give me a help? The intuition says that the gradient is zero at the origin.
 A: You know, do you not, that $|x|^p= (x_1^2+ x_2^2+ \cdots+ x_n^2)^{p/2}$?
The gradient of a function of $n$ variables is the $n$-dimensional vector whose $i$th component is the derivative of that function with respect to $x_i$.
A: Let $u$ be a unit vector.  You can compute $D_uf(0)$, the derivative of $f$ in the direction of $u$ at the origin, by the definition:
$$
D_uf(0) = \lim_{h\to 0} \frac{f(hv)}{h} 
= \lim_{h\to0} \frac{|hv|^p}{h}
= \lim_{h\to0} \frac{|h|^p|v|^p}{h}
= \lim_{h\to0} h^{p-1} |v|
= 0
$$
The last limit is zero since $p\geq 2 \implies p-1 \geq 1$.
The gradient vector $\nabla f(0)$ is characterized by the property that $\nabla f(0) \cdot u = D_u f(0)$ for all unit vectors $u$.  But the only vector whose dot product with every unit vector is zero, is the zero vector.  Thus $\nabla f(0) = 0$.
A: Clearly $f(0,0)= 0.$ We have 
$$[f(0,h)-f(0,0)]/(h - 0) = f(0,h)/h = |h|^p/h.$$
The last term $\to 0$ as $h\to 0.$ Thus $\partial f /\partial x_2(0,0) = 0.$ Same for $\partial f /\partial x_1.$ Thus $\nabla f(0,0) = (0,0).$
