# Multi-variable limit existence

I am bit confused with multi-variable limit existence and particularly path test taking example $$\lim\limits_{(x,y) \to (0,0)} {x+y \over x-y} = {0\over0} (indeterminate )$$ when approach the limit along y = mx $$\lim\limits_{(x,y) \to (0,0)} {x+y \over x-y} = {x+mx\over x-mx} = {1+m\over1-m}$$ so the limit value varies by changing m value and it is no longer dependent on $$x$$&$$y$$ values and if we take the limit when approach the limit along y = mx $$\lim\limits_{(x,y) \to (0,1)} {x+y \over x-y} = {x+mx\over x-mx} = {1+m\over1-m}$$ it is also not exist but if we directly substituted in the original limit with (0,1) we will find that the limit exists and equals to $$-1$$ can anyone explain where is my mistake or explain why this happened ?

• You are taking limit as $x$and $y$ both tend to $0$. So the value when $x=0$ and $y=1$ has no connection with the limit. – Kavi Rama Murthy Mar 25 '20 at 9:57
• I can't say what you're asking is clear. – Allawonder Mar 25 '20 at 10:04
• How many of the lines $y=mx$ pass through the point $(0,1)$? – Brian Moehring Mar 25 '20 at 10:10
• @BrianMoehring thanks for your comment , Peter explained it to me in his answer below – Jimmy Morgan Mar 25 '20 at 11:26

Since $$y=mx$$ contains the point $$(0,0)$$ for every real $$m$$, this approach shows that we have different limits depending on the direction we approach the point $$(0,0)$$. Hence the limit for $$(x,y)\rightarrow (0,0)$$ does not exist.
• We have a rational function in $x$ and $y$ and it is well known that the such a function is continous wherever it is defined. Hence the limit for $(x,y)->(0,1)$ exists and it does not matter how we approach this point. For this point, $y=mx$ is not valid, as other users have already mentioned. – Peter Mar 25 '20 at 11:00
You have made a mistake in subtitution. when $$(x,y) → (0,0)$$ you cannot choose y = mx path. This path should be used for $$(x,y) → (0,0)$$. So $$y = mx$$ is not a suitable path.