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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 ?

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  • $\begingroup$ 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. $\endgroup$ – Kavi Rama Murthy Mar 25 '20 at 9:57
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    $\begingroup$ I can't say what you're asking is clear. $\endgroup$ – Allawonder Mar 25 '20 at 10:04
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    $\begingroup$ How many of the lines $y=mx$ pass through the point $(0,1)$? $\endgroup$ – Brian Moehring Mar 25 '20 at 10:10
  • $\begingroup$ @BrianMoehring thanks for your comment , Peter explained it to me in his answer below $\endgroup$ – Jimmy Morgan Mar 25 '20 at 11:26
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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.

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  • $\begingroup$ yes i understand that but this isn't the case when (x,y) -> (0,1) it still depends on value of m (the direction we approach point (0,1) in this case ) although the limit exist in this case $\endgroup$ – Jimmy Morgan Mar 25 '20 at 10:56
  • $\begingroup$ 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. $\endgroup$ – Peter Mar 25 '20 at 11:00
  • $\begingroup$ Ok , but one last question why y = mx isn't valid ? Isn't the limit supposed to exist when we get to it through all paths ? $\endgroup$ – Jimmy Morgan Mar 25 '20 at 11:07
  • $\begingroup$ All paths that approach the given point. $\endgroup$ – Peter Mar 25 '20 at 11:08
  • $\begingroup$ Oh i get it , thanks $\endgroup$ – Jimmy Morgan Mar 25 '20 at 11:20
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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.

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