Question

How do you prove that the limit of `(x^3 - 3) = 5` as x approaches 2 using the epsilon delta proof?

Answer 1

For any `epsilon >0`, we fix `delta_epsilon` such that:

`0 < delta_epsilon < min(1, epsilon/19)`

Explanation:

`f(x) = (x^3-3)` is a polynomial function therefore continuous in all of `RR`, so we know that:

`lim_(x->2) f(x) = f(2) = 5`

To give a formal proof we analyze the expression `abs (f(x) -5)` around `x=2` expressing `x` as `2+Deltax`:

`abs (f(x) -5) = abs ((2+Deltax)^3-3-5) =abs (8+12(Deltax)+6(Deltax)^2+(Deltax)^3-8)=abs(12(Deltax)+6(Deltax)^2+(Deltax)^3)`

Now chosen any `epsilon >0`, we fix `delta_epsilon` such that:

`0 < delta_epsilon < min(1, epsilon/19)`

and as for `x in (2-delta_epsilon,2+delta_epsilon)` we have `|Deltax| < delta_epsilon`

`abs (f(x) -5) = abs(12(Deltax)+6(Deltax)^2+(Deltax)^3)<= 12abs(Deltax)+6abs(Deltax)^2+abs(Deltax)^3< 12delta_epsilon+6delta_epsilon^2+delta_epsilon^3`

now, as `delta_epsilon < 1` we have `delta_epsilon^n < delta_epsilon`, so:

`abs (f(x) -5) < 12delta_epsilon+6delta_epsilon^2+delta_epsilon^3 < 12delta_epsilon+6delta_epsilon+delta_epsilon < 19delta_epsilon < epsilon`

And the limit is proved.