# 2 path test for non-existence

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?

• 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. Mar 14, 2019 at 17:54
• @Yanko , why can't i use x = 0? Mar 14, 2019 at 17:58
• @Yanko so i have to use paths x=1 and y= -1? Mar 14, 2019 at 17:59
• 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. Mar 14, 2019 at 19:09

## 1 Answer

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.

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