# Epsilon delta limit problem

## 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-xol<$$\delta$$ the inequality lf(x)-Ll<$$\epsilon$$ holds
f(x)=x2
L=3
x0=-2
$$\epsilon$$=0.5

## 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 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 $$\delta$$=$$\sqrt{}4.5$$-2 as an answer

## The Attempt at a Solution

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LCKurtz
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## 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 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 $$\delta$$=$$\sqrt{}4.5$$-2 as an answer
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?