Find the limit or show that it does not exist:
$\lim _{\left(x,\:y\right)\to \left(1,\:-1\right)}\left(\frac{xy+1}{x^2-y^2}\right)$

For this question i have used 2 different paths:

Path 1: $x=0$, where,
$\lim _{\left(y\right)\to \left(-1\right)}\left(\frac{1}{-y^2}\right) = -1$

Path 2: $y=0$, where,
$\lim _{\left(x\right)\to \left(1\right)}\left(\frac{1}{x^2}\right) = 1$

This yielded 2 different limits, hence the limit does not exist. Is this the right approach, and if so, can I use any path to my liking?

  • $\begingroup$ There's a problem with your path. You must get $x$ and $y$ to $1$ and $-1$. You can't put $x=0$, you can try $x=1$ instead. $\endgroup$
    – Yanko
    Mar 14, 2019 at 17:54
  • $\begingroup$ @Yanko , why can't i use x = 0? $\endgroup$ Mar 14, 2019 at 17:58
  • $\begingroup$ @Yanko so i have to use paths x=1 and y= -1? $\endgroup$ Mar 14, 2019 at 17:59
  • $\begingroup$ You don't have to. But you can. You must choose paths that tends to $(1,-1)$. The path $(0,y)$ as $y\rightarrow -1$ goes to $(0,-1)$ so you can't use it. $\endgroup$
    – Yanko
    Mar 14, 2019 at 19:09

1 Answer 1


You're right. The sequence doesn't converge, however your reasoning is wrong.

The fact that $\lim_{(x,y)\rightarrow(1,-1)} f(x,y)$ exists tells you nothing about how $f$ behaves around $(0,-1)$ and therefore you putting $x=0$ makes no sense.

If $(x,y)\rightarrow (1,-1)$ then so is $(1,y)$. Therefore if (by contradiction) $\lim_{(x,y)\rightarrow(1,-1)} f(x,y)$ exists so does $\lim_{y\rightarrow -1} f(1,y)$ and both limits must be equal.

$f(1,y) =\frac{y+1}{1-y^2} = \frac{1+y}{(1-y)(1+y)} = \frac{1}{1-y}$ whenever $y\not = -1$. Therefore $\lim_{y\rightarrow -1} f(1,y) = \frac{1}{2}$.

Similar calculation would yield that $\lim_{x\rightarrow 1} f(x,-1) = -\frac{1}{2}$.

However the existence of $\lim_{(x,y)\rightarrow(1,-1)} f(x,y)$ implies that

$$-\frac{1}{2} =\lim_{x\rightarrow 1} f(x,-1) = \lim_{(x,y)\rightarrow(1,-1)} f(x,y) = \lim_{y\rightarrow -1} f(1,y)= \frac{1}{2}$$ which is a contradiction.

  • $\begingroup$ Thank you, your explanation makes sense:) @Yanko $\endgroup$ Mar 15, 2019 at 18:34

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