$\lim$ $(x,y)$ approaches $(1,0)$ for $(x^2-y^2-1)/(x^2 +y^2 -1)$?

Question

If I substitute $x=1$ and $y=0$ directly,I get $0/0$ so I subtitute $y=x-1$ and found the limit is $1$?But apparently it's undefined so what else do I substitute with to get another limit that is not $1$?

Answer 1

From your work, substituting $y=x-1$, we have $$\lim_{(x,y)\rightarrow(1,0)} \frac{x^2-y^2-1}{x^2+y^2-1}=\lim_{x\rightarrow 1}\frac{2x-2}{2x^2-2x}=\lim_{x\rightarrow 1}\frac{1}{x}=1$$

On the other hand, substituting $x=1$, we have $$\lim_{(x,y)\rightarrow(1,0)} \frac{x^2-y^2-1}{x^2+y^2-1}=\lim_{y\rightarrow 0}\frac{-y^2}{y^2}=-1$$

It may seem like my substitution of "$x=1$" is invalid (since we are told that we cannot substitute like this in 2-D limits). But since we are proving in contradiction, with the fact that "if the limit exists, then any (relevant) substitution will give the same result", its contrapositive is that "if two (relevant) substitutions give different result, the limit does not exist". So I think I have finished the proof.

Answer 2

If the limit exists it should b consistant

$\lim_{y\to 0}(\lim_{x\to 1} \frac {x^2 -y^2 - 1}{x^2 + y^2 -1} ) =$

$\lim_{y\to 0} \frac {-y^2}{y^2} = \lim_{y\to 0} -1 = -1$.

But $\lim_{x\to 1}(\lim_{y\to 0} \frac {x^2 -y^2 - 1}{x^2 + y^2 -1} ) =$

But $\lim_{x\to 1}\frac {x^2 -1}{x^2 -1} = \lim_{x\to 1} 1 = 1$.

If we to $x = t; y = 1-t$ then

$\lim_{(x,y)\to (1,0)}\frac {x^2 -y^2 - 1}{x^2 + y^2 -1} = \lim_{t\to 1} \frac {t^2 - (1 - 2t + t^2) - 1}{t^2 + (1 - 2t + t^2)-1}=$

$\lim_{t\to 1}\frac { 2t -2}{2t^2-2t}=\lim_{t\to 1}\frac 1{t} = 1$.

And if $y = t$ and $x= 1-t$ then

$\lim_{(x,y)\to (1,0)}\frac {x^2 -y^2 - 1}{x^2 + y^2 -1} = \lim_{t\to 0} \frac {(1-t)^2 -t^2 -1}{(1-t)^2 +2t +1}=$

$\lim_{t\to 0}\frac { -2t}{2+t^2}=0$.

No consistant limit exists.