Question

How do you use the epsilon delta definition to prove that the limit of `(1+x)/(4+x)=1/2` as `x->2`?

Answer 1

See below.

Explanation:

Given a positive `epsilon`, let `delta = min{1, 14epsilon}`. Note that `delta` is positive.

For any `x` with `0 < abs(x-2) < delta`, we have

`abs(x-2) < 1`, which is equivalent to `-1 < x-2 < 1`, and this entails that `-5 < x+4 < 7`, so we will have `-10 < 2(x+4) < 14` and, finally we are assured that `abs(2(x+4)) < 14`

So for any `x` with `0 < abs(x-2) < delta`, we have

`abs((1+x)/(4+x) - 1/2) = abs((2+2x-4-x)/(2(x+4)))`

`= abs((x-2)/(2(x+4))`

`< (abs(x-2))/14`

`< delta/14`

`< (14epsilon)/14`

`= epsilon`

We have shown that for any `epsilon > 0` there is a `delta > 0` such that

for any `x` with `0 < abs(x-2) < delta`

`abs((1+x)/(4+x) - 1/2)< epsilon`