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### A Two-Variable Delta-Epsilon Proof

```
Date: 02/05/2001 at 11:50:44
From: Calculus fan
Subject: A problem in limits of two variable function

Dear Doctor Math,

I would really like it if you would help me prove the following
statement using the delta-epsilon definition of the limit.

0 =    lim    (x*sqrt(1+y) + y*sqrt(1+x))
(x,y)->(0,0)

Thanks.
```

```
Date: 02/07/2001 at 21:31:33
From: Doctor Fenton
Subject: Re: A problem in limits of two variable function

Dear Fan,

Thanks for writing to Dr. Math. Your problem is this: you are
assuming that someone has given you a small positive number named
epsilon (which I am going to write as just "e"), and you must
determine a radius delta ("d") for a small disk centered on (0,0)
such that if (x,y) is a point in the disk, then

|x*sqrt(1+y) + y*sqrt(1+x)| < e

Well, if (x,y) is in a disk of radius d, then

x^2 + y^2 <= d

and in particular,

|x|  = sqrt(x^2)
<= sqrt(x^2+y^2)
<= d, and similarly,
|y| <= d

We know from the Triangle Inequality that

|x*sqrt(1+y) + y*sqrt(1+x)| <= |x|*sqrt(1+y) + |y|*sqrt(1+x)

Since we are only concerned with small disks, we can let d<=1. We
know that when d<=1, then |x|<=1 and |y|<=1. So

sqrt(1+x) <= sqrt(2) < 2
sqrt(1+y) <= sqrt(2) < 2

Thus, as long as d<=1,

|x*sqrt(1+y) + y*sqrt(1+x)| < |x|*2 + |y|*2

We know that |x|<=d and |y|<=d, so

|x*sqrt(1+y) + y*sqrt(1+x)| < |x|*2 + |y|*2

< 2*d + 2*d

< 4*d

So, how can we choose d to make sure the quantity on the left is less
than e? Can you finish it from here?

If you have further questions, please write again.

- Doctor Fenton, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Calculus

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