# Epsilon-delta proof

• August 15th 2010, 11:26 AM
usagi_killer
Epsilon-delta proof
Use an $\epsilon - \delta$ proof to show that $\lim_{(x,y) \to (0,0)} (y-x)\log_e(x^2+y^2) = 0$

So we need to prove that if $0 < \sqrt{x^2+y^2} < \delta$ then $|(y-x)\log_e(x^2+y^2)| < \epsilon$

However I do not know where to go from here, I can not seem to manipulate the $x^2+y^2$ and relate it with $\log_e(x^2+y^2)$ with any inequalities.

Thank you all!
• August 15th 2010, 03:03 PM
Vlasev
I think you should use start with the fact that [LaTeX ERROR: Convert failed] .
• August 16th 2010, 03:15 AM
usagi_killer
Thanks for that, I've played around with that however I still can not seem to get it :(
• August 16th 2010, 04:10 AM
HallsofIvy
Try going to polar coordinates. With $x= r cos(\theta)$ and $y= r sin(\theta)$, $x^2+ y^2= r^2$ so that $ln(x^2+ y^2)= ln(r^2)= 2 ln(r)$, $(y- x)ln(x^2+ y^2)= (rcos(\theta)- rsin(theta))ln(r^2)= 2r ln(r)(cos(\theta)- sin(\theta))$. Now, just take some upper bound on $cos(\theta)- sin(\theta)$ (it certainly can't be larger than 2) and you have reduced to a one-variable, r, problem.
• August 16th 2010, 04:16 AM
thorin
Hi,

$|y-x| \leq |x|+|y| \leq 2 \sqrt{x^2+y^2}$

So

$|(x-y)log(x^2+y^2)| \leq 4 \sqrt{x^2+y^2} log (\sqrt{x^2+y^2} )
$

now, we can use the fact that x.ln(x) has 0 for limit in 0 :
$\forall \epsilon>0 \ \exists \delta>0 \ |X| \leq \delta ==> |x.ln(x)| \leq \epsilon
$
with
$X=\sqrt{x^2+y^2}
$
• August 16th 2010, 04:49 AM
Vlasev
HallsOfIvy, please refer to my post in http://www.mathhelpforum.com/math-he...es-153726.html to see an example where your method fails. It fails because there are other ways to approach the limit point, not just straight lines. There is no way to reduce the 2 dimensional problem to a 1 dimensional problem unless it has circular symmetry around the point.
• August 16th 2010, 06:11 AM
usagi_killer
Quote:

Originally Posted by thorin
Hi,

$|y-x| \leq |x|+|y| \leq 2 \sqrt{x^2+y^2}$

So

$|(x-y)log(x^2+y^2)| \leq 4 \sqrt{x^2+y^2} log (\sqrt{x^2+y^2} )
$

now, we can use the fact that x.ln(x) has 0 for limit in 0 :
$\forall \epsilon>0 \ \exists \delta>0 \ |X| \leq \delta ==> |x.ln(x)| \leq \epsilon
$
with
$X=\sqrt{x^2+y^2}
$

Hi thanks for the help, however I am unclear how you got the first line of your working.

Thanks!
• August 16th 2010, 06:20 AM
Vlasev
Quote:

Originally Posted by usagi_killer
Hi thanks for the help, however I am unclear how you got the first line of your working.

Thanks!

The left one is the triangle inequality.

$\displaystyle |x-y| = |x+(-y)| \leq |x|+|-y| = |x|+|y|$

Using that you have

$\displaystyle |x|+|y| = \sqrt{x^2}+\sqrt{y^2} \leq \sqrt{x^2+y^2}+\sqrt{x^2+y^2} = 2\sqrt{x^2+y^2}$
• August 16th 2010, 06:21 AM
usagi_killer
Oh right, thanks, I get it now!