Using epsilon-delta show that limit x->2 x^3 = 8?

Given ε > 0, we need to find δ > 0 such that 0 < |x - 2| < δ ==> |x³ - 8| < ε.

To this end, note that

|x³ - 8|

= |((x - 2) + 2)³ - 8|, by adding 0 cunningly

= |((x - 2)³ + 3 * 2(x - 2)² + 3 * 4(x - 2) + 8) - 8|, by multiplication/Binomial Theorem

= |(x - 2)³ + 6(x - 2)² + 12(x - 2)|

≤ |(x - 2)³| + |6(x - 2)²| + |12(x - 2)|, by the triangle inequality

= |x - 2|³ + 6|x - 2|² + 12|x - 2|

< |x - 2| + 6|x - 2| + 12|x - 2|, assuming that |x - 2| < 1 (to be used for δ)

= 19|x - 2|.

So given ε > 0, let δ = min{1, ε/19}.

Then, 0 < |x - 2| < δ ==> |x³ - 8| < 19|x - 2| < 19(ε/19) = ε, as required.

I hope this helps!

without the cunning...| x³ - 8 | = | x - 2 | | x² + 2x + 4 |...and if δ ≤ 1 then | x | ≤ 3===> | x² + 2x + 4 | ≤ 19 ===> | x³ - 8 | ≤ 19 | x - 2 |

so choose δ ≤ max { ε / 19 , 1 }