Determine $\lim_{(x, y)\to (0, 0)}\frac{-x+y+1}{x^2-y^2}$ In wolframalpha I tried to calculate 
$$\lim_{(x, y)\to (0, 0)}\dfrac{-x+y+1}{x^2-y^2}$$
and it returns :
(limit does not exist, is path dependent, or cannot be determined)
can't we say directly that the limit is $\dfrac{1}{0}=\infty$ so it does not exist ? does this mean that the limit is path independent since for any path we will find the limit equals to $\infty$? and what about the third possibility of "cannot be determined" does this mean that it may or may not exist but the software of wolframalpha is unable to decide? thank you for your help!
 A: The answer that Wolfram Alpha is giving you is that the limit cannot be found. 
That could be because of several reasons.


*

*The "limit" might fail to be finite

*The "limit" might depend on how you arrive

*Wolfram might not know how, or it might be impossible to find, the "limit".


Case 1:
Limits don't exist if the don't tend towards a finite value, e.g.
$$\lim_{(x,y) \to (0,0)} \frac{1}{x^2+y^2} = \infty$$
Case 2: 
Some limits give finite values, but the value you get depends on how you arrive at $(0,0)$, e.g.
$$\lim_{(x,y) \to (0,0)} \frac{x}{\sqrt{x^2+y^2}}$$
If we put $x=r\cos\theta$ and $y=r\sin\theta$ then we get
$$\lim_{r \to 0^+} \frac{r\cos\theta}{\sqrt{r^2\cos^2\theta+r^2\sin^2\theta}} = \cos\theta.$$
Even though $|\cos\theta| \le 1$, the limit depends on the direction and so is not unique.
Case 3: The third choice is that Wolfram has no idea.
Your limit is a bit of a mixture between Case 1 and Case 2. It does not tend to a finite limit, but it also tends to $+\infty$ and $-\infty$, depending on how you arrive at the origin.
A: On the angles $x^2>y^2$ the limit is $+\infty$, while on the angles $x^2<y^2$ the limit is $-\infty$. One could say that in such cases the limit is $\infty$ (without sign). Just a matter of definitions...
A: Consider the paths over the axis: $y=0$ and $x<0$ AND $x=0$ and $y>0$.
