I'd like to know what's the derivative of $f(x,y)=\cos{\sqrt{x^2+y^2}}$ at $(0, 0)$.
WolframAlpha says it is indeterminate.
However, if we apply the definition, we can actually evaluate it to zero:
$$ f_x(0,0) = \lim_{h\to 0} \dfrac{f(h,0) - f(0,0)}{h} = \lim_{h\to0} \dfrac{\cos{|h|}-1}{h} = \lim_{h\to0} \dfrac{\cos{h}-1}{h} = 0 \\ f_y(0,0) = \lim_{k\to 0} \dfrac{f(0,k) - f(0,0)}{k} = \lim_{k\to0} \dfrac{\cos{|k|}-1}{k} = \lim_{k\to0} \dfrac{\cos{k}-1}{k} = 0 $$
Aside question: if Wolfram is wrong, how often it happens in your experience? Does it ever happen?
Interestingly, this function is not only differentiable at $(0,0)$ (with the official definition), but in fact $C^1$. If one makes the standard mistake (which I've seen instructors make in teaching single-variable calculus) of blithely saying $f'(0)=\lim_{x\to 0} f'(x)$, it does work out nicely here. Note that \begin{align*} \lim_{(x,y)\to (0,0)} \frac{\partial f}{\partial x}(x,y) &= \lim_{(x,y)\to (0,0)} \left(\frac x{\sqrt{x^2+y^2}}\right)\left(-\sin\sqrt{x^2+y^2}\right) \\ &= \lim_{(x,y)\to (0,0)} -x\left(\frac{\sin\sqrt{x^2+y^2}}{\sqrt{x^2+y^2}}\right) = 0\cdot 1 = 0. \end{align*} Nothing whatsoever indeterminate. I think @Dunkelheit's comment is on point. Since I do not use Wolfram ordinarily to answer such questions, I had not previously bumped into this phenomenon.