Existence of derivative at the origin I'm supposed to find for which values of $p$ and $q$ the following function is differentiable at the origin:
$$   
f(x,y,z) = 
     \begin{cases}
       \dfrac{(x^2y^2)^p(1-\cos(z))^q}{x^2+y^2+z^2} & \text{if } (x,y,z) \neq (0,0,0) \\
        0 & \text{if } (x,y,z) = (0,0,0)\\
      \end{cases}
$$
I've made some progress using the Taylor expansion of the cosine about $0$, but couldn't get far even doing so. Could anyone share a solution?
 A: Hint: You can check that the partial derivatives of $f$ at $(0,0,0)$ are all $0.$ Thus $Df(0,0,0)$ exists iff
$$f(x,y,z) = f(0,0,0) + 0 +o((x^2+y^2+z^2)^{1/2})= o((x^2+y^2+z^2)^{1/2})$$
as $(x,y,z) \to (0,0,0).$ In other words, $Df(0,0,0)$ exists iff
$$\lim_{(x,y,z) \to (0,0,0)} \frac{f(x,y,z)}{(x^2+y^2+z^2)^{1/2}} = 0.$$
A: Here's a full solution as you asked, following previously given hints. Notice that using Taylor's expansion as you suggested gives us the numerator $(x^{2p}y^{2p}z^{2q})/2$, so that
the limit we want to be zero becomes
$$\lim_{(x,y,z) \to (0,0,0)} \frac{f(x,y,z)}{(x^2+y^2+z^2)^{1/2}} = \lim_{(x,y,z) \to (0,0,0)} \frac{x^{2p}y^{2p}z^{2q}}{2(x^2+y^2+z^2)^{3/2}}.$$
The trick lies in using the next three inequalities
$$ x^2 \leq x^{2}+y^{2}+z^{2}$$
$$ y^2 \leq x^{2}+y^{2}+z^{2}$$
$$ z^2 \leq x^{2}+y^{2}+z^{2}$$
to write
$$\lim_{(x,y,z) \to (0,0,0)} \frac{x^{2p}y^{2p}z^{2q}}{2(x^2+y^2+z^2)^{3/2}} \leq \lim_{(x,y,z) \to (0,0,0)} \frac{(x^2+y^2+z^2)^{2p+q}}{2(x^2+y^2+z^2)^{3/2}}$$
from where it follows the limit is zero whenever $2p +q$ is strictly greater than $3/2$.
