1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Epsilon-delta proof

  1. Mar 22, 2006 #1
    I'm working on an Epsilon-delta proof for:

    [tex]\lim_{(x,y)\rightarrow (1,1)} x^2 + xy + y = 3[/tex]

    I've tired a couple of different approaches but am not getting anywhere (I'm terrible at these). Any suggestions would be appreciated. Thanks.

  2. jcsd
  3. Mar 22, 2006 #2
    Would you show us what you already tried?
  4. Mar 23, 2006 #3


    User Avatar
    Homework Helper

    Start off with the definition:

    [tex]\lim_{(x,y)\rightarrow (1,1)} x^2 + xy + y = 3 \mbox{ if, and only if }[/tex]
    [tex]\mbox{for every }\epsilon >0,\mbox{ there exists a }\delta >0 \mbox{ such that } 0<\sqrt{(x-1)^2+(y-1)^2}<\delta \mbox{ implies that }|x^2 + xy + y - 3| <\epsilon[/tex]​
  5. Mar 23, 2006 #4


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Prove that the sum of two continuous functions is itself a continuous function.
  6. Mar 23, 2006 #5

    Thanks benorin. I was trying with [tex]0<\sqrt{x^2+y^2}<\delta[/tex]. I think I can get it now.

  7. Mar 23, 2006 #6


    User Avatar
    Staff Emeritus
    Science Advisor

    By the way, while [itex]\sqrt{(x_1-x_0)^2+ (y_1-y_0)^2}[/itex] is the "standard metric" it would be equivalent to show the "for every [itex]\epsilon>0[/itex] there exist [itex]\delta> 0[/itex] such that max (|x-1|,|y-1|)< [itex]\delta[/itex] implies that |x2+ xy+ y- 3|<[itex]\epsilon[/itex] and might be simpler. Of course, your teacher might require that you be able to show that they are equivalent!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Epsilon-delta proof
  1. Epsilon-Delta Proof (Replies: 3)

  2. Delta epsilon proof (Replies: 5)