Prove the given problem that involves limits

[chwala]

Homework Statement

This is my own question (set by me).

$f(x)=x^3+2x^2-4x-8$ Prove that

$$\lim_{x→ 2} f(x) =0 $$

Relevant Equations

uniform continuity.

I am self-learning analysis.

My steps are as follows,
For any $ε>0$, there is a $\delta>0$ such that, $|(x^3+2x^2-4x-8) -0|<ε$ when $0 < |x-2|<\delta$

Let $\delta≤1$ then $1<x<3, x≠2$.
$|(x^3-2x^2-4x-8) -0|=|(x-2)(x+2)^2|=|x-2||(x+2)^2| <\delta |(x+2)^2|<19\delta$
Taking $\delta$ as $1$ or $\dfrac{ε}{19}$ whichever is smaller. Then we have, $|(x^3+2x^2-4x-8) -0|<ε$ whenever $0 < |x-2|<\delta$.

insight is welcome.

 

[PeroK]

Where did the factor of 19 come from?

 

[chwala]

I noted that,

$|x^2+2x+4|$ is bounded by $[7,19]$.

 

[PeroK]

I noted that,

$|x^2+2x+4|$ is bounded by $[7,19]$.

$(x+2)^2 = x^2 + 4x + 4$, which is clearly bounded by 25 when $1 < x < 3$.

 

[chwala]

$(x+2)^2 = x^2 + 4x + 4$, which is clearly bounded by 25 when $1 < x < 3$.

Ah... i made a mistake. Noted.


Then the last part changes to,

Let $\delta≤1$ then $1<x<3, x≠2$.
$|(x^3-2x^2-4x-8) -0|=|(x-2)(x+2)^2|=|x-2||(x+2)^2| <\delta |(x+2)^2|<25\delta$
Taking $\delta$ as $1$ or $\dfrac{ε}{25}$ whichever is smaller. Then we have, $|(x^3+2x^2-4x-8) -0|<ε$ whenever $0 < |x-2|<\delta$.