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/
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