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:

$$\lim_{(h,k)\to(0,0)} \frac{[f(1+h,1+k)-f(1,1)-h(\partial_xf(1,1))-k(\partial_yf (1,1))]}{\sqrt{h^2+k^2}} $$


$$\frac{\partial f}{\partial x}(1,1) = 1, \quad \frac{\partial f}{\partial y}(1,1) = 1$$

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!

  • $\begingroup$ What function?? $\endgroup$ Jun 29, 2018 at 15:20
  • 3
    $\begingroup$ Please, use MathJax for math formatting. It makes things readable, especially long expressions like that limit of yours. $\endgroup$
    – Arthur
    Jun 29, 2018 at 15:22
  • $\begingroup$ alright, can't right now but if this remains unanswered by tonight I will make better-formatted question. Sorry guys! Edit: Thank you Lorenzo!! $\endgroup$
    – Rick
    Jun 29, 2018 at 15:42
  • 1
    $\begingroup$ @Rick In the case of a multivariate function there are signifincant differences between "partially differentiable" and "total differentiable". I assume you mean the latter. $\endgroup$
    – Peter
    Jun 29, 2018 at 15:55

1 Answer 1


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 $h= k$

$$\lim_{(h,k)\to(0,0)} \frac{[f(1+h,1+k)-f(1,1)-h(\partial_xf(1,1))-k(\partial_yf (1,1))]}{ [(h^2 + k^2)^{1/2}]}=\lim_{(h,k)\to(0,0)} \frac{h+2-2-2h}{h\sqrt 2}=-\frac1{\sqrt 2}$$

  • $\begingroup$ Thank you so much. My problem was I was substituting (1+h,1+k) into x + y instead of into x + 1, I guess I need more coffee. $\endgroup$
    – Rick
    Jun 29, 2018 at 16:45
  • $\begingroup$ @Rick You are welcome! Bye $\endgroup$
    – user
    Jun 29, 2018 at 16:45

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