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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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2  
do you mean $\log\frac{1+y^2}{x^2+xy}$? –  user127.0.0.1 Dec 1 '12 at 18:30
1  
yes that is what i meant –  question Dec 1 '12 at 18:33
    
Welcome to math.SE. For some basic information about writing math at this site see e.g. here, here and here. –  Américo Tavares Dec 1 '12 at 18:33
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What for? $\lim_{(x,y)\rightarrow(1,0)} \ln\left(\frac{1+y^2}{x^2+xy}\right) = \ln\left(\frac{1+0^2}{1^2+1\cdot 0}\right) = \ln(1) = 0$ –  user127.0.0.1 Dec 1 '12 at 18:42
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to make sure the limit always give the same value no matter the direction and thus exist –  question Dec 1 '12 at 18:44

1 Answer 1

up vote 4 down vote accepted

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+1 –  question Dec 1 '12 at 18:39
    
i get ln(1/2) when y = x-1 –  question Dec 1 '12 at 18:44
    
yeah when i do the hospital rule it gives me 1/2 –  question Dec 1 '12 at 18:45
    
nah how come i get ln(1/2) when y = x-1 and 0 when y = 0? –  question Dec 1 '12 at 18:47
    
i need to approach (1,0) from all direction and get the same value, but i don't, so it musn't exist, right? –  question Dec 1 '12 at 18:50

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