How to prove a limit exists using delta and epsilon?

Question

I am having issues writing a proof using deltas and epsilons for this limit: I basically do not how to construct a proof after finding a limit. My main problem with problems like that arise from the proof. I took the the path of y=x and y =-x to come up with L= 0. $$\lim_{(x,y) \to (0,0)} \frac{xy}{\sqrt{x^2+y^2}}$$

Thanks, Mo.

I would try going into polar coordinates, writing $x = r\cos\theta$ and $y = r\sin\theta$. — user164587, Sep 13 '14 at 01:43

score 7 · Accepted Answer · edited Apr 13 '17 at 12:21

7

Related techniques: (I). Here is a how.

$$ \Bigg| \frac{xy}{\sqrt{x^2+y^2}}-0 \Bigg| = \frac{|x|\,|y|}{\sqrt{x^2+y^2}} \leq \frac{\sqrt{x^2+y^2}\,\sqrt{x^2+y^2}}{\sqrt{x^2+y^2}} = \sqrt{x^2+y^2} < \epsilon = \delta. $$

Note:

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

edited Apr 13 '17 at 12:21

Community

1

answered Sep 13 '14 at 02:00

Mhenni Benghorbal

47,431

1

Short and sweet! Good work! – user175623 Sep 13 '14 at 02:02
@user175623: You are welcome. – Mhenni Benghorbal Sep 13 '14 at 02:03

score 0 · Answer 2 · answered Sep 13 '14 at 01:46

0

What do you think the limit is? If I give you an $\epsilon$ that is positive, you need to find a $\delta$ so that over the whole square $-\delta \lt x,y \lt \delta$ the expression is within $\epsilon$ of the limit. It seems that $x=\pm y$ will yield the extreme values. What are your issues?

answered Sep 13 '14 at 01:46

Ross Millikan

374,822

The limit = 0, but I don't know how to prove it. – user175623 Sep 13 '14 at 01:48
user164587's comment is quite useful. You will get a function of $r$ times some trig functions in $\theta$. The trig functions are easy to bound. – Ross Millikan Sep 13 '14 at 01:51

How to prove a limit exists using delta and epsilon?

2 Answers2