# Homework Help: Find limit L. Then find δ > 0 such that |f(x) - L| < 0.01 (f(x) = x^2 -3)

1. Oct 14, 2012

1. The problem statement, all variables and given/known data
Find limit L. Then find δ > 0 such that |f(x) - L| < 0.01

Limit when x approaches 2 of x^2 -3

2. Relevant equations
0 < |x-c| < δ
0 < |f(x) - L| < ε

3. The attempt at a solution
function is continuous at D = ℝ so limit is f(2) = 1

0 < |x-2| < δ and

0 < |x^2 -3 -1| < 0.01 ⇔ 0 < |x^2 -4| <0.01 ⇔ 0 < |x + 2|*|x-2| < 0.01 which is good because:
0 < |x-2| = 0.01 / |x+2|

I've seen the solution and i see that i'm supposed to assume a range for x (like (1,3) ). I can imagine that because the function isn't linear a range has to be assumed.

They say that assuming this range gives δ = 0.01 / 5 = 0.002 witch seems to be the smallest of 0.01 / 3 and 0.01 / 5. That makes sense to me.

But why this chosen range? Why does this range apply to ε = 0.01? Choosing a different range gives a different δ.

Who can help me with this i'm really trying to understand this.

2. Oct 14, 2012

### tiny-tim

exactly!

δ depends on ε

(we could write it δ(ε) )

these proofs all involve showing that whatever ε we choose, we can always find a δ

(usually δ gets smaller and smaller, just like ε)

3. Oct 14, 2012

Hey Tiny Tim

If i Would express δ(ε) I'd say δ(ε) = ε / |x+2|. Calculating δ with this function is possible for D = ℝ (for both ε and x) so a logical conclusion is that the limit exists that's what you mean right?

but ε = 0.01 is given to find the appropriate δ. Something has to be assumed for x in δ(ε) = ε / |x+2|. They assume (1,3) and find δ = 1/5*ε or δ = 1/3*ε and therefore conclude that δ = 1/5 * 0.01 = 0.002. That i don't get. What's the relation between the chosen range (1,3) an ε = 0.01? Why is that valid?

Last edited: Oct 14, 2012