1
$\begingroup$

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?

$\endgroup$
4
  • $\begingroup$ 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

0
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Thank you, your explanation makes sense:) @Yanko $\endgroup$ Mar 15, 2019 at 18:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.