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Epsilon delta limit problem

  • Thread starter miglo
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Homework Statement


find an open interval about x0 on which the inequality lf(x)-Ll<[tex]\epsilon[/tex] holds. Then give a value for [tex]\delta[/tex]>0 such that for all x satisfying 0<lx-xol<[tex]\delta[/tex] the inequality lf(x)-Ll<[tex]\epsilon[/tex] holds
f(x)=x2
L=3
x0=-2
[tex]\epsilon[/tex]=0.5


Homework Equations





The Attempt at a Solution


0<lx+2l<[tex]\delta[/tex] [tex]\Rightarrow[/tex] lx2-4l<0.5
-0.5<x2-4<0.5
3.5<x2<4.5
[tex]\sqrt{}3.5[/tex]<x<[tex]\sqrt{}4.5[/tex]

so i thought my interval about x0 would be ([tex]\sqrt{}3.5[/tex],[tex]\sqrt{}4.5[/tex]) but my book says its the negative of both those numbers, so im thinking its because x approaches -2, then my interval would be on the negative side of the x-axis?
also i dont know how my book got [tex]\delta[/tex]=[tex]\sqrt{}4.5[/tex]-2 as an answer
please help me with these limit problems, im really confused

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
LCKurtz
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The Attempt at a Solution


0<lx+2l<[tex]\delta[/tex] [tex]\Rightarrow[/tex] lx2-4l<0.5
-0.5<x2-4<0.5
3.5<x2<4.5
[tex]\sqrt{}3.5[/tex]<x<[tex]\sqrt{}4.5[/tex]

so i thought my interval about x0 would be ([tex]\sqrt{}3.5[/tex],[tex]\sqrt{}4.5[/tex]) but my book says its the negative of both those numbers, so im thinking its because x approaches -2, then my interval would be on the negative side of the x-axis?
also i dont know how my book got [tex]\delta[/tex]=[tex]\sqrt{}4.5[/tex]-2 as an answer
please help me with these limit problems, im really confused
When you solve for x values at the end points of your inqualities where you have x2 = 3.5 and x2= 4.5, you should get two solutions:

[tex]x =\pm \sqrt{3.5}\hbox{ and }x=\pm\sqrt{4.5}[/tex]

The values near x = -2 are the negative ones. You should draw this parabola and draw the lines y = 3.5 and y = 4.5 on your graph. You will see that x values near +2 and near -2 give y values between 3.5 and 4.5.

What remains is for you to find the correct interval for x nearby -2 and figure out what δ works. Note that |x + 2| < δ describes a symmetric interval about -2 and your interval isn't symmetric.

Edit: Isn't your L a typo and should be 4?
 

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