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I got $$ \lim_{(x,y)\to (1,0)} \ln\left(\frac{1+{y^2}}{{x^2} +xy}\right) $$

and I get

  • $\lim = 0,$ when $y = 0,$
  • $\lim = 0,$ when $y = x,$

but when $y = -x$ I get undefined. So the limit doesn't exist?

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    do you mean $\log\frac{1+y^2}{x^2+xy}$?2012-12-01
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    yes that is what i meant2012-12-01
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    Welcome to math.SE. For some basic information about writing math at this site see e.g. [here](http://meta.math.stackexchange.com/questions/5020/), [here](http://meta.stackexchange.com/a/70559/155238) and [here](http://meta.math.stackexchange.com/questions/1773/).2012-12-01
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    Then .. just insert (1,0) into your expression. There is no reason for a limit-calculation.2012-12-01
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    why? shouldn't i calculate the limit?2012-12-01
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    What for? $\lim_{(x,y)\rightarrow(1,0)} \ln\left(\frac{1+y^2}{x^2+xy}\right) = \ln\left(\frac{1+0^2}{1^2+1\cdot 0}\right) = \ln(1) = 0$2012-12-01
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    to make sure the limit always give the same value no matter the direction and thus exist2012-12-01
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    The only interesting limit you could calculate are those, where $(x,y)$ goes to some point with $x^*=-y^*$. In all other cases you can calculate the value of ln() directly.2012-12-01
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    i don't understand why it doesn't work.2012-12-01
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    what do you mean?2012-12-01
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    @question the fact that all limits in directions exist doesn't mean that the limit exists! Consider a set $\{(x,y):(|x|-1)^2+y^2<1\}\cup\{(x,y):x=0\}$ and the characteristic function of this set. Then the linear "cuts" of this function going through $(0,0)$ are all continous at this point, but the function itself is not continuous!2012-12-01
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    if they have all the same values it exists2012-12-01

1 Answers 1

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You need to get straight about what is approaching what:

$$ \lim_{(x,y)\to (1,0)} \ln\left(\frac{1+{y^2}}{{x^2} +xy}\right) $$

means you need to take the limit of the point $(x, y)\in \mathbb{R}^2$ as $(x, y)$ approaches the point $(1, 0)$, i.e. as both $x\to 1$ AND $y \to 0$.

$$ \lim_{(x,y)\to (1,0)} \ln\left(\frac{1+{y^2}}{{x^2} +xy}\right) = \ln\left(\frac{1 + 0^2}{1^2 + 1\cdot 0}\right) = \ln\left(\frac{1}{1}\right) = \ln 1 = 0$$.

So no worries.

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    ah yeah y=-x it should have been y=-x+12012-12-01
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    i get ln(1/2) when y = x-12012-12-01
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    yeah when i do the hospital rule it gives me 1/22012-12-01
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    nah how come i get ln(1/2) when y = x-1 and 0 when y = 0?2012-12-01
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    i need to approach (1,0) from all direction and get the same value, but i don't, so it musn't exist, right?2012-12-01
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    when y = x-1 = 0, x = 12012-12-01
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    i know, but i don't get ln(1) when i do the hospital rule after substituting y by x-1.2012-12-01
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    You are not taking the limit $ \lim_{(x,y)\to (1,0)} \left(\frac{1+{y^2}}{{x^2} +xy}\right)$. You are taking the limit $ \lim_{(x,y)\to (1,0)} \ln\left(\frac{1+{y^2}}{{x^2} +xy}\right)$. l'hospital is not needed, nor appropriate here.2012-12-01
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    why? so you just replace x by 1?2012-12-01
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    Yes, just replace x by 1, and y by 0.2012-12-01
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    thank you, can you tell me in 1 short sentence why the hospital rule isn't appropriate though?2012-12-01
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    @question Read l'Hopital's rule. It only applies when the limit of the top and bottom are both 0 or both infinity. We don't have either in this case so it doesn't apply. We don't even have the limit of a fraction. You have the limit of the natural log of a fraction, which is not the same thing.2012-12-01
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    question: @Graphth answered your question as best as I could.2012-12-01