Why Does My Solution Differ From the Textbook in This Epsilon-Delta Problem?

[miglo]

Homework Statement

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

Homework Equations

The Attempt at a Solution

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

so i thought my interval about x₀ would be (\sqrt{}3.5,\sqrt{}4.5) but my book says its the negative of both those numbers, so I am thinking its because x approaches -2, then my interval would be on the negative side of the x-axis?
also i don't know how my book got \delta=\sqrt{}4.5-2 as an answer
please help me with these limit problems, I am really confused

 

[LCKurtz]

The Attempt at a Solution

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

so i thought my interval about x₀ would be (\sqrt{}3.5,\sqrt{}4.5) but my book says its the negative of both those numbers, so I am thinking its because x approaches -2, then my interval would be on the negative side of the x-axis?
also i don't know how my book got \delta=\sqrt{}4.5-2 as an answer
please help me with these limit problems, I am 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:

x =\pm \sqrt{3.5}\hbox{ and }x=\pm\sqrt{4.5}

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?