# Thread: Epsilon-delta proof

1. ## Epsilon-delta proof

Use an $\displaystyle \epsilon - \delta$ proof to show that $\displaystyle \lim_{(x,y) \to (0,0)} (y-x)\log_e(x^2+y^2) = 0$

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

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

Thank you all!

2. I think you should use start with the fact that $\displaystyle u > \ln(u)$.

3. Thanks for that, I've played around with that however I still can not seem to get it

4. Try going to polar coordinates. With $\displaystyle x= r cos(\theta)$ and $\displaystyle y= r sin(\theta)$, $\displaystyle x^2+ y^2= r^2$ so that $\displaystyle ln(x^2+ y^2)= ln(r^2)= 2 ln(r)$, $\displaystyle (y- x)ln(x^2+ y^2)= (rcos(\theta)- rsin(theta))ln(r^2)= 2r ln(r)(cos(\theta)- sin(\theta))$. Now, just take some upper bound on $\displaystyle cos(\theta)- sin(\theta)$ (it certainly can't be larger than 2) and you have reduced to a one-variable, r, problem.

5. Hi,

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

So

$\displaystyle |(x-y)log(x^2+y^2)| \leq 4 \sqrt{x^2+y^2} log (\sqrt{x^2+y^2} )$
now, we can use the fact that x.ln(x) has 0 for limit in 0 :
$\displaystyle \forall \epsilon>0 \ \exists \delta>0 \ |X| \leq \delta ==> |x.ln(x)| \leq \epsilon$with
$\displaystyle X=\sqrt{x^2+y^2}$

6. 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.

7. Originally Posted by thorin
Hi,

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

So

$\displaystyle |(x-y)log(x^2+y^2)| \leq 4 \sqrt{x^2+y^2} log (\sqrt{x^2+y^2} )$
now, we can use the fact that x.ln(x) has 0 for limit in 0 :
$\displaystyle \forall \epsilon>0 \ \exists \delta>0 \ |X| \leq \delta ==> |x.ln(x)| \leq \epsilon$with
$\displaystyle X=\sqrt{x^2+y^2}$

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

Thanks!

8. 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.

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

Using that you have

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

9. Oh right, thanks, I get it now!