Help with epsilon delta proof

Question

$x^3$ is near $27$ when $x$ is near $3$ but $x$ is not equal to $3$.

So I have

$$0<|x-3|<\delta \implies 3-\delta<x<3+\delta$$

$$|x^3-27|<\epsilon \implies|(x-3)(x^2+3x+3^2)|<\epsilon$$

$$=(x^2+3x+9)|(x-3)|<\epsilon \implies28|(x-3)|<\epsilon$$

$$=|(x-3)|<\epsilon/28$$

How do I prove $x^2+3x+9$ is less than $28$?

My attempt

$|x-3|<\delta$

so $$|x^2+3x+9|=|x^2+3x-18+27|\leq|x^2+3x-18|+27 \\ =|(x-3)(x+6)|+27 =|(x-3)(x-3+9)|+27$$

I'm not really sure where to go from here.

Answer 1

You can continue by saying that $|(x-3)(x-3+9)|+27\le|x-3|(|x-3|+9)+27<\delta(\delta+9)+27$,

so now you just have to make sure that $\delta(\delta+9)\le1$ by, say, choosing $\delta$ so that it satisfies $\delta\le\frac{1}{18}$

since then $\delta^2\le\frac{1}{2}$ and $9\delta\le\frac{1}{2}$

(along with any other conditions required to ensure that $|x^3-27|<\epsilon$).

Answer 2

There's a safer way to think. Let $\epsilon > 0$, and suppose that $0 < |x - 3| < \delta$. let's find $\delta$. Notice: $$|x| - 3 < |x-3| < \delta \implies |x| < \delta + 3.$$ Once we find $\delta$, any other value $\delta'$ less than this $\delta$ will also work, so we impose that $\delta \leq 1$. If the value of $\delta$ we find is greater than $1$, if we choose anything less than $1$, it will also work. We have: $$|x^3 - 27| = |(x - 3)(x^2+3x+9)| = |x^2+3x+9||x-3|.$$ We use the triangle inequality, and the hypothesis $|x-3| < \delta $ to get: $$|x^3-27| \leq (x^2+3|x|+9)\delta.$$ Now, we use that $|x| < 3+\delta < 3+1 = 4$. Hence: $$|x^3-27| < (16+12+9)\delta = 37\delta$$ I want $37\delta < \epsilon$, so $\delta < \epsilon/37$.

The crux here is that this argument only holds if $\delta < 1$ too. So, our final answer is that: given $\epsilon > 0$, if $\delta = \min\{1, \epsilon/37\}$, then: $$0<|x-3|<\delta \implies |x^3 - 27| < \epsilon.$$

You might find my answer here helpful. I give the general strategy for dealing with epsilon-delta proofs for limits with polynomials.