Please verify my epsilon - delta proof $\lim_{x\to 2}⁡(x^3 )=8,$ and $0 < x < 4$

Question

I am concern about the delta calculation given $0 < x < 4$. I believe it works, but I am not 100 percent sure as I am new to proofs.

Consider the function $f(x) = x^3$ for $x ∈ ℝ$ and $0 < x < 4$. Prove that $$\lim _{ x\rightarrow 2 }{ { x }^{ 3 }=8 } $$

Proof: Let $\varepsilon > 0$ be given, then let $\delta = \min {(\varepsilon/19,1)}$. Now suppose $|x -2| < \delta = \min {(\varepsilon /19,1)}$. Then $|x^3 - 8| = |x - 2||x^2 + 2x + 4|< |x - 2|<19$ $(|x^2 + 2x + 4| < 19~\text{if}~|x - 2| < 1)< \varepsilon /19 $ $(|x - 2| < \varepsilon /19 )= \varepsilon $.

Thus this $\delta$ makes $|x^3 - 8| < \varepsilon $ whenever $0 < |x - 2| < \delta$. Therefore, it follows that $\lim _{ x\rightarrow 2 } x^3 = 8$.

Answer 1

Need some improvement in presentation.

If $|x-2|< \delta = \min(\epsilon/19,1),$

then we have $x \in (1,3)$ and hence $|x^2+2x+4|<19.$

Hence, for $|x-2|< \delta$, we have $$|x^3-8|=|x-2||x^2+2x+4|< (\epsilon/19)(19)=\epsilon $$

Remark:

The line $|x-2||x^2+2x+4|<\left|x-2\right|$ is wrong in your proof.

Answer 2

Perhaps a simpler road is this.

let $\epsilon \in R^+$ $$|x^3-8|<\epsilon$$ now substitute $x \to t+2$, and the condition becomes: $$ |t^3+6t^2+12t|<\epsilon $$ if we choose $|t| < \min\{\epsilon/19,1\}$ the condition is respected and by back substitution: $$|x-2| < \min\{\epsilon/19,1\}\equiv\delta \implies |x^3-8|<\epsilon$$ which is the definition of limit.