Recent content by la_med12

Ah.. I guess I was looking at it for so long that I overlooked the obvious. Thanks for the help!

I'm learning how to find delta from a particular epsilon. I'm not understanding a step in the solution for the problem listed below: Here's the problem: lim(x→3)x^2=9 Solution: |x^2-9|<.05 -.05<x^2-9<.05 2.9916..<x<3.0083 2.9916..-3<x-3<3.0083..-3 -.0083..<x-3<.0083.. delta=.0083...

Thanks for the check. Cheers!

1. I'm pretty sure I've got this right.. just needing a check. Cheers! 2. x^2+2xy-y^2+x^2=2 3. 2x+2x(dy/dx)+2y-2y(dy/dx)+2x=0 2x(dy/dx)-2y(dy/dx)=-4x-2y dy/dx(2x-2y)=-4x-2y dy/dx=(-4x-2y)/(2x-2y)