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 MurthyCommented Mar 25, 2020 at 9:57
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1$\begingroup$ I can't say what you're asking is clear. $\endgroup$– AllawonderCommented Mar 25, 2020 at 10:04
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1$\begingroup$ How many of the lines $y=mx$ pass through the point $(0,1)$? $\endgroup$– Brian MoehringCommented Mar 25, 2020 at 10:10
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$\begingroup$ @BrianMoehring thanks for your comment , Peter explained it to me in his answer below $\endgroup$– Jimmy MorganCommented Mar 25, 2020 at 11:26
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2 Answers
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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 $\endgroup$ Commented Mar 25, 2020 at 10:56
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$\begingroup$ 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. $\endgroup$– PeterCommented Mar 25, 2020 at 11:00
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$\begingroup$ 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 ? $\endgroup$ Commented Mar 25, 2020 at 11:07
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$\begingroup$ All paths that approach the given point. $\endgroup$– PeterCommented Mar 25, 2020 at 11:08
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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.
