This was the question of a test. My question is if my attempt to solve it is correct, and if it is, why is it correct.

$$\lim_{(x,y) \to (0,0), x+y \neq 0}{\frac{\ln(1-x-y)}{x+y} } $$

My attempt:

Let $\xi = -x-y $. Then $\xi \to 0$ whenever $(x,y) \to (0,0)$ and $x+y \neq 0 \iff \xi \neq 0$. (Is it then correct to say that the previous limit exists and is equal to the following iff the following exists? And why?):

$$\lim_{\xi \to 0, \xi \neq 0}{\frac{\ln(1+\xi )}{-\xi}}$$

If it is correct, then the limit exists and is $-1$. If it is correct, why is it correct?

  • $\begingroup$ If you can show from the $\epsilon-\delta$-definition (take $x^2+y^2<\delta$) that the function is continuous, then it means that the limit exists and is unique from any path. $\endgroup$ – D.B. Apr 6 '19 at 17:20
  • $\begingroup$ @D.B. the function is elementary, so I can assume it is continuous in it's domain.. however it is undefined in the point $(0,0)$. $\endgroup$ – RUBEN GONÇALO MOROUÇO Apr 6 '19 at 17:30
  • $\begingroup$ Right, it may not be continuous at $0$. $\endgroup$ – D.B. Apr 6 '19 at 17:33
  • $\begingroup$ I haven't had Calculus III since 2008, but from what I recall from 2 variable functions like this, you can try to graph them to visually find the answer. Just go to wolframalpha.com and input "3d plot Log[1-x-y]/(x+y)" (without quotes) into the search bar. You will find that as x approaches 0, z = f(x,y) approaches 1. As y approaches 0, z approaches negative infinity. So the limit probably does not exist. $\endgroup$ – Christopher Mowla Apr 6 '19 at 21:04
  • $\begingroup$ @ChristopherMowla curiously if I ask wolfram to evaluate that limit it says it's -1.. But wolfram alpha is not reliable for multivariable limits. It often lies. Although so, it would be strange if it didn't exist, since the question explicity tells me to calculate it's value. And I don't think they'd put limits that are too hard to solve in the test. $\endgroup$ – RUBEN GONÇALO MOROUÇO Apr 6 '19 at 21:13

We want to show that $\forall \epsilon \gt 0 : \exists \delta \gt 0 : ||(x,y)||< \delta \implies |\frac{\ln(1-x-y)}{x+y}+1| \lt \epsilon$.

Fix $\epsilon \gt 0$. We know that $\lim_{\phi \to 0} \frac{ln(1+\phi)}{-\phi} = -1$. So there's $\delta_1 \gt 0$ such that $|\phi| \lt\delta_1 \implies |\frac{ln(1+\phi)}{-\phi}+1| \lt \epsilon$. Let $\delta = \delta_1$ and $\xi = -x-y$. Suppose $||(x,y)|| \lt \delta $. Since all norms in $R^n$ are equivalent we can use the norm of the sum. Then:

$||(x,y)|| = |x|+|y| \geq |x+y| = |\xi|$. Since $|\xi| < \delta_1 $, then $|\frac{ln(1+\xi)}{-\xi}+1| < \epsilon$ therefore $|\frac{ln(1-x-y)}{x+y}+1| < \epsilon $. $\square$


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