I have spent over two hours trying to understand why this function is not differentiable at $(1,1)$! $$f(x,y)=\begin{cases}x+y & x\ne y\\x+1 &x=y \end{cases}$$ Supposedly we ought to prove it through using the following equation:
$$\lim_{(h,k)\to(0,0)} \frac{[f(1+h,1+k)-f(1,1)-h(\partial_xf(1,1))-k(\partial_yf (1,1))]}{\sqrt{h^2+k^2}} $$
with:
$$\frac{\partial f}{\partial x}(1,1) = 1, \quad \frac{\partial f}{\partial y}(1,1) = 1$$
the limit is $0$ when $h \neq k$
But supposedly when $h = k$ the limit is different from $0$ which proves it is not differentiable at $(1,1)$ but no matter what I do I can't seem to get a result different from $0$ when $h = k$
Any help would be much appreciated!
You are right, the function is not differentiable, by definition, since the limit doesn't exist when $(x,y)\to (1,1)$.
Indeed for $h\neq k$
$$\lim_{(h,k)\to(0,0)} \frac{[f(1+h,1+k)-f(1,1)-h(\partial_xf(1,1))-k(\partial_yf (1,1))]}{ [(h^2 + k^2)^{1/2}]}=\lim_{(h,k)\to(0,0)} \frac{h+k+2-2-h-k}{ [(h^2 + k^2)^{1/2}]} =0$$
while for $h= k$
$$\lim_{(h,k)\to(0,0)} \frac{[f(1+h,1+k)-f(1,1)-h(\partial_xf(1,1))-k(\partial_yf (1,1))]}{ [(h^2 + k^2)^{1/2}]}=\lim_{(h,k)\to(0,0)} \frac{h+2-2-2h}{h\sqrt 2}=-\frac1{\sqrt 2}$$