Question

How do you prove a limit of (x^2+3) as x approaches 1 equal 4? Thanks

Please show the process using Epsilon-Delta proof. Thanks again

Answer 1

Evaluate the difference:

`abs(x^2+3-4) = abs (x^2-1)`

`abs(x^2+3-4) = abs (x-1)abs(x+1)`

Let now `x=1+xi`, then:

`abs(x^2+3-4) = abs xi abs ( 2 +xi)`

Given any number `epsilon > 0` choose:

`delta_epsilon < min(1,epsilon/3)`

For `x in (1-delta_epsilon,1+delta_epsilon)` we have that:

`abs xi < delta_epsilon`

Now, as `delta_epsilon < 1` we have that:

`abs(2+xi) <= 2+abs xi < 2+1 = 3`

and then:

`abs(x^2+3-4) = abs xi abs ( 2 +xi) < 3abs(xi)`

and because `delta_epsilon < epsilon/3` we have that:

`abs(x^2+3-4) < 3abs(xi) < 3 * epsilon/3 = epsilon`

We can conclude that for every `epsilon` if we choose `delta_epsilon < min(1,epsilon/3)` we have that:

`x in (1-delta_epsilon,1+delta_epsilon) => abs(x^2+3-4) < epsilon`