Question

How do you prove that the limit of `x^2 = 9` as x approaches 3 using the epsilon delta proof?

Answer 1

See below.

Explanation:

Prove `lim_(x->3)x^2=9`.

The epsilon delta proof for limits is easier understood when one is familiar with the definitions of the terms involved. Most useful will be the definition of the limit of a function.

`lim_(x->c)f(x)=L`

Definition:

• Let `f: D->RR`
• Let `c` be an accumulation point of `D`
• We say that the limit of `f(x)` at `c` is the real number `L` provided that:

for every `epsilon>0`, there exists `delta>0` such that `abs(f(x)-L)< epsilon` whenever `0< abs(x-c)< delta`.

In our case:

• `f(x)=x^2`
• `c=3`
• `L=9`

We then need to find an expression for `delta` so that the definition, and consequently the proof, holds. In other words, we need:

`abs(x^2-9)< epsilon" "` when `" "0< abs(x-3)< delta`

So:

`abs(x^2-9)< epsilon`

`=>abs((x-3)(x+3))< epsilon`

Since the concept of the limit only applies when `x` is close to `a`, we will need to restrict `x` so that it is at most `1` away from `a`, or: `abs(x-a) <1`. For us, that is `abs(x-3)< 1`.

`=>-1< (x-3) <1`

`=>2< x <4`

`=>5< (x+3) <7`

`=> (x+3) <7`

We are looking at the maximum values for both cases.

So we know that we must have:

`abs((x-3)(x+3))< 7delta=epsilon`

`=>delta=epsilon/7`

Then our two restrictions are:

`abs(x-3)<1` and `abs(x-3)< delta/7`

We want a `delta` which will satisfy both inequalities, so we'll end up taking the minimum value for `epsilon` between `1` and `epsilon/7`.

Note that all of the above steps come from the definition of the limit of a function as provided above.

Proof. Let `epsilon>0` be arbitrary and let `delta=min{1,(epsilon)/7}`. If `abs(x-3)< delta`, then `abs(x-3) <1`; hence:

`-1< x-3 <1`

`=>2< x< 4`

`=>5< x+ 3< 7`

`=>abs( x+3)< 7`

Also, if `abs(x-3)< delta`, then `abs(x-3)< epsilon/7`, so:

`abs(x^2-9)=abs((x-3)(x+3))`

`<7*abs(x-3)`

`<=7*epsilon/7=epsilon" " square`

---

Another way to go about working out the scratch work is as follows:

We know we need `abs(x^2-9)< epsilon`

`=>abs((x-3)(x+3))< epsilon`

`=>abs(x-3)< epsilon/(abs(x+3))`

`=>delta < epsilon/(abs(x+3))`

`abs(x-3)<1`

`=>-1< abs(x-3) < 1`

`=>2< x < 4`

`=> 5< (x+3) <7`

So, for

`abs(x-3)< epsilon/(abs(x+3))`

we can see that the RHS is at a minimum when `(x+3)` is at a maximum, i.e. the LHS decreases as the RHS increases. The maximum of `x+3` was found above to be `7`.

`:. delta=epsilon/7`