I'm having difficulties in understanding proofs of limits.This is all new to me. The teacher doesn't cover all of the material in the lectures.

Could someone please explain the following example? Where did theinterval (1,3)come from and thex + 2 < 5?

Prove: lim_{x→2}x^{2}= 4

We have to show that for each ε > 0, there's a ∂ > 0 such that |x^{2}- 4| < ε whenever 0 < |x - 2| < ∂.

|x^{2}- 4| = |x - 2|*|x + 2|

The example then reports that for all x in interval (1,3), x + 2 < 5 and that therefore |x + 2| < 5. (How do we know that? Where does this come from?)

Then we set ∂ to be the minimum of ε/5 and 1. (Ditto... Where does this come from? What does it mean to be the "minimum"? What's the max.?)

Whenever 0 < |x - 2| < ∂, we have |x^{2}- 4| = |x - 2|*|x + 2| < (ε/5)(5) = ε

For x-values within ∂ of 2, f(x) values are within ε of 4.

And here's a homework problem:

Prove: lim_{x→9}(x-5)^{1/2}= 2.

We have to show that for all x, 0<|x-9|<∂, so that also |(x-5)^{1/2}-2| < ε.

First, solve inequality |(x-5)^{1/2}-2| < ε.

This gets: 9 - 4ε + ε^{2}< x < 9 + 4ε + ε^{2}

So far so good.Now what?

We have the interval (9 - 4ε + ε^{2}, 9 + 4ε + ε^{2})

Now find value for ∂ > 0.

∂min = {|9 -(9 - 4ε + ε^{2})|, |9 - (9 + 4ε + ε^{2})|} (What does minimum mean in this context? Why does ∂ equal it?)

∂min = {|- 4ε + ε^{2})|, | 4ε + ε^{2})|}

∂ = ??? (Don't know.)

Thanks for any help! Sorry if this seems/is dumb. I haven't got my head around the formal definition yet.