Prove the given problem that involves limits

chwala
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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.
 
Last edited:
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Where did the factor of 19 come from?
 
I noted that,

##|x^2+2x+4|## is bounded by ##[7,19]##.
 
chwala said:
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##.
 
PeroK said:
##(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##.
 
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