Question

How do you find the Limit of `2x+5` as `x->-3` and then use the epsilon delta definition to prove that the limit is L?

Answer 1

If `x` is very close to `-3`,

then `2x` is very close to `2(-3) = -6`

and `2x+5` is very close to `-6+5 = -1`.

So `lim_(xrarr-3)(2x+5) = -1`

Proof:

Given a positive number, `epsilon`, let `delta = epsi/2`.

Then for every `x` with `0 < abs (x-(-3)) < delta`, we have `abs(x+3) < epsi/2`.

Therefore `abs((2x+5)-(-1))` which is equal to `abs(2x+6)` is equal to `2abs(x+3)` is less than `2(delta)` which is `2(epsi/2)` or just `epsi`.

Writing it in mathematics we write:

For `0 < abs (x-(-3)) < delta`, we have

`abs((2x+5)-(-1)) = abs(2x+6)`

`= 2abs(x+3)`

`< 2(delta)`

`= 2(epsi/2) = epsi`.

So, by the definition of limit, `lim_(xrarr-3)(2x+5) = -1`