Q) Prove that $ \displaystyle \lim_{x \to 2} x^3 = 8$.
This is how I approached it:
$ \implies |x^3 - 2^3|<\epsilon$
$ \implies |(x-2)(x+2)^2|< \epsilon$
$ \implies |x-2||x+2|^2<\epsilon.$
then I tried to find a bound of $|x+2|^2$,by restricting the x at most away from $1$ i.e $|x-2|<1$;
$-1<x-2<-1$
$3<x+2<5$
$3<|x+2|<5$
$9<|x+2|^2<25$, so $|x+2|^2$ was bounded above by $25$, so,
$|x-2| < \frac{\epsilon}{25}$,
so $\delta$ = min{$1, \frac{\epsilon}{25}$}
Let $\delta$ be $1$, so $1 < \frac{\epsilon}{25}$, $25 < \epsilon$, |x-2| < 1 ==> $|x-2| < 1 < \frac{\epsilon}{25}$
==> $|x-2| < \frac{\epsilon}{25}$
==> $|x-2||x+2|^2 < \frac{\epsilon}{25} * |x+2|^2$
==> $|(x-2)(x+2)^2 | < \frac{\epsilon}{25} *25 $
==> $|x^3 - 2^3| < \epsilon$
==> $|f(x) - L| < \epsilon.$
Is this correct? Do correct me if I was wrong at any point.
It seems your factorization of $x^3-8$ is wrong.
$$x^3-8 = (x-2)(x^2+2x+4)$$
Th rest of the procedure is the same as you have done
$$-1 < x-2 < 1$$ Implies
$$1<x<3$$
Or
$$7=1^2+2(1)+4<x^2+2x+4<3^2+2(3)+4=19$$
This means that we can choose $\delta = \min \left(1, \frac{\epsilon}{19} \right)$ such that
$$|f(x)-f(2)| = |x^3 - 2^3| = |x-2| \cdot |x^2+2x+4| \le \delta \cdot (x^2+2x+4) \le \frac{\epsilon}{19} \cdot 19 = \epsilon $$
Which completes the proof.