We know that linear approximation applies if a function is differentiable; that is, if for $f(x,y)$: $$\lim_{(h,k)\to (0,0)} \frac{f(a+h,b+k)-f(a,b)-hf_x(a,b)-kf_y(a,b)}{\sqrt{h^2+k^2}}=0$$
Since this is often difficult to compute, the following condition is usually used:
If the partial derivatives $f_x$ and $f_y$ exist near $(a,b)$ and are continuous at $(a,b)$, then $f$ is differentiable at $(a,b)$.
Now, I am trying to find whether linear approximation applies for $f(x,y)=x \cos y$ at the point $(x_0,y_0)=(0,0)$. We find that $f_x(x,y)=\cos y$ and $f_y(x,y)=-x\sin y$. Using the above criterion, $f$ should be differentiable at $(0,0)$, since $f_x(0,0)=1$ and $f_y(0,0)=0$, which are both continuous.
This is all fine, but when I try to use the definition of differentiability, I encounter a problem:
$$\lim_{(h,k)\to (0,0)} \frac{f(h,k)-f(0,0)-hf_x(0,0)-kf_y(0,0)}{\sqrt{h^2+k^2}}=\lim_{(h,k)\to (0,0)}\frac{h(\cos k-1)}{\sqrt{h^2+k^2}}$$ I have not managed to compute this limit manually, but Wolfram|Alpha claims that it does not exist. My question is: how can this limit not exist (implying the function is not differentiable) when the other condition was satisfied. Did I make a mistake somewhere in the process, or is it not supposed to agree in this case? Thank you for any assistance!
Use polar coordinates ($h=r\cos\phi$ and $k=r\sin\phi$). It's completely straightforward.
$$\lim_{(h,k)\to (0,0)}\frac{h(\cos k-1)}{\sqrt{h^2+k^2}}=\lim_{r\rightarrow 0} \frac{r\cos\phi(\cos(r\sin\phi)-1))}{r}=\lim_{r\rightarrow 0} \cos\phi(\cos(r\sin\phi)-1))=0$$
Also please note than the above "limit condition" is weaker than continuity of partials. There are differentiable functions with discontinuous partials!