Question

How do you find the Limit of `x+3` as `x->2` and then use the epsilon delta definition to prove that the limit is L?

Answer 1

See below.

Explanation:

To evaluate, we can simply take the limit by substituting `2` in for `x`, and we then have `(2+3)=5`.

Now, prove `lim_(x->2)x+3=5`.

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+3`
• `c=2`
• `L=5`

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+3-5)< epsilon" "` when `" "0< abs(x-2)< delta`

So:

`abs(x-2)< 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-2)< 1`.

In this case, we have `x-2` for both cases, so we need not evaluate both.

Then we know that we must have:

`abs(x-2)< delta=epsilon`

`=>delta=epsilon`

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

Since we have `delta=epsilon` and both cases were the same, the proof will be pretty short.

Proof. Let `epsilon>0` be arbitrary and let `delta=epsilon`. If `-1 < abs(x-2)< delta`, then `abs((x+3)-2)=abs(x-2) <1`; hence:

`abs(x-2) <1`

`<= delta=epsilon" " square`