Finding $\displaystyle\lim_{x \to 2} (x^3 - x) = 6$ using epsilon-delta.

Question

Find $$\displaystyle\lim_{x \to 2} (x^3 - x) = 6$$ using epsilon-delta.

I have attempted the following:

Given $ε>0$, the exists a particular chosen $δ$.

Then whenever $0<|x-2|<δ$, $|x^3 - x - 6| = |x(x^2 - 1) - 6| = |x(x-1)(x+1) - 6|$

I do not know how to continue from here.

Is it possible to do the following?

$|x ((x-2)+1) ((x-2)+3) - 6| ≤ |x (|x-2|+1) (|x-2|+1) - 6|$?

Thanks and sorry for the confusion.

Answer 1

We have $|x^3-x-6|=|x-2||x^2+2x+3|$. For $x$ near $2$ we have $1 \le x \le 3$, hence

$|x^3-x-6| \le |x-2|(9+6+3)=18|x-2|$.

Your turn !