How do you use the epsilon delta definition to prove that the limit of #x^3=27# as #x->3#?

To prove that the limit of x^3 as x approaches 3 is equal to 27 using the epsilon-delta definition, we proceed as follows:

Let ε be a positive number. We need to find a corresponding δ such that if 0 < |x - 3| < δ, then |x^3 - 27| < ε.

First, we can rewrite |x^3 - 27| as |(x - 3)(x^2 + 3x + 9)|.

Now, we want to find an upper bound for |x^2 + 3x + 9| in terms of δ. We can do this by considering the interval (2, 4) for x, since we want x to approach 3.

For any x in this interval, we have |x^2 + 3x + 9| ≤ (4^2 + 3(4) + 9) = 37.

Now, let's choose δ = min(1, ε/37).

If 0 < |x - 3| < δ, then it follows that |x^3 - 27| = |(x - 3)(x^2 + 3x + 9)| < δ * 37 ≤ (ε/37) * 37 = ε.

Therefore, we have shown that if 0 < |x - 3| < δ, then |x^3 - 27| < ε, which satisfies the epsilon-delta definition of a limit.

Hence, the limit of x^3 as x approaches 3 is indeed equal to 27.