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Prove that $lim_x to 2 x^3 = 8$ by using epsilon-delta

2

1

$begingroup$

Prove that $$lim_x to 2 x^3 = 8$$

My attempt,

Given $epsilon>0$, $exists space delta>0$ such that if $$|x^3-8|<epsilon space textif space 0<|x-2|<delta$$

$$|(x-2)(x^2+2x+4)|<epsilon$$

I'm stuck here. Hope someone could continue the solution and explain it for me. Thanks in advance.

limits proof-verification epsilon-delta

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asked Jan 5 '18 at 12:53

MathxxMathxx

3,41811544

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add a comment | 

2

1

$begingroup$

Prove that $$lim_x to 2 x^3 = 8$$

My attempt,

Given $epsilon>0$, $exists space delta>0$ such that if $$|x^3-8|<epsilon space textif space 0<|x-2|<delta$$

$$|(x-2)(x^2+2x+4)|<epsilon$$

I'm stuck here. Hope someone could continue the solution and explain it for me. Thanks in advance.

limits proof-verification epsilon-delta

share|cite|improve this question

asked Jan 5 '18 at 12:53

MathxxMathxx

3,41811544

$endgroup$

add a comment | 

$begingroup$

Prove that $$lim_x to 2 x^3 = 8$$

My attempt,

Given $epsilon>0$, $exists space delta>0$ such that if $$|x^3-8|<epsilon space textif space 0<|x-2|<delta$$

$$|(x-2)(x^2+2x+4)|<epsilon$$

I'm stuck here. Hope someone could continue the solution and explain it for me. Thanks in advance.

limits proof-verification epsilon-delta

share|cite|improve this question

asked Jan 5 '18 at 12:53

MathxxMathxx

3,41811544

$endgroup$

share|cite|improve this question

asked Jan 5 '18 at 12:53

MathxxMathxx

3,41811544

share|cite|improve this question

asked Jan 5 '18 at 12:53

MathxxMathxx

3,41811544

asked Jan 5 '18 at 12:53

MathxxMathxx

3,41811544

asked Jan 5 '18 at 12:53

MathxxMathxx

3,41811544

2 Answers
2

active

oldest

votes

0

$begingroup$

Note that

$$x^2 + 2x + 4 = (x-2)^2 + 6(x-2) + 12$$

so
beginalign|(x-2)(x^2 + 2x + 4)| &= |(x-2)((x-2)^2 + 6(x-2) + 12)|\
&le |x-2|(|x-2|^2 + 6|x-2| + 12)
< delta(delta^2 + 6 delta + 12)
endalign

For $varepsilon > 0$ you would take $delta < minleftfracvarepsilon19, 1right$ because then $|x-2| < delta$ implies:

$$|x^3 - 8| le delta(delta^2 + 6delta + 12) < fracvarepsilon19 (1 + 6 + 12) = varepsilon$$

share|cite|improve this answer

answered Jan 5 '18 at 12:58

mechanodroidmechanodroid

28.9k62648

$endgroup$

add a comment | 

0

$begingroup$

For $xin [1,3]quad |(x-2)(x^2+2x+4)|le 19|x-2|<epsilon$

share|cite|improve this answer

answered Jan 5 '18 at 13:05

StuStu

1,1951414

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add a comment | 

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0

$begingroup$

Note that

$$x^2 + 2x + 4 = (x-2)^2 + 6(x-2) + 12$$

so
beginalign|(x-2)(x^2 + 2x + 4)| &= |(x-2)((x-2)^2 + 6(x-2) + 12)|\
&le |x-2|(|x-2|^2 + 6|x-2| + 12)
< delta(delta^2 + 6 delta + 12)
endalign

For $varepsilon > 0$ you would take $delta < minleftfracvarepsilon19, 1right$ because then $|x-2| < delta$ implies:

$$|x^3 - 8| le delta(delta^2 + 6delta + 12) < fracvarepsilon19 (1 + 6 + 12) = varepsilon$$

share|cite|improve this answer

answered Jan 5 '18 at 12:58

mechanodroidmechanodroid

28.9k62648

$endgroup$

add a comment | 

0

$begingroup$

Note that

$$x^2 + 2x + 4 = (x-2)^2 + 6(x-2) + 12$$

so
beginalign|(x-2)(x^2 + 2x + 4)| &= |(x-2)((x-2)^2 + 6(x-2) + 12)|\
&le |x-2|(|x-2|^2 + 6|x-2| + 12)
< delta(delta^2 + 6 delta + 12)
endalign

For $varepsilon > 0$ you would take $delta < minleftfracvarepsilon19, 1right$ because then $|x-2| < delta$ implies:

$$|x^3 - 8| le delta(delta^2 + 6delta + 12) < fracvarepsilon19 (1 + 6 + 12) = varepsilon$$

share|cite|improve this answer

answered Jan 5 '18 at 12:58

mechanodroidmechanodroid

28.9k62648

$endgroup$

add a comment | 

$begingroup$

Note that

$$x^2 + 2x + 4 = (x-2)^2 + 6(x-2) + 12$$

so
beginalign|(x-2)(x^2 + 2x + 4)| &= |(x-2)((x-2)^2 + 6(x-2) + 12)|\
&le |x-2|(|x-2|^2 + 6|x-2| + 12)
< delta(delta^2 + 6 delta + 12)
endalign

For $varepsilon > 0$ you would take $delta < minleftfracvarepsilon19, 1right$ because then $|x-2| < delta$ implies:

$$|x^3 - 8| le delta(delta^2 + 6delta + 12) < fracvarepsilon19 (1 + 6 + 12) = varepsilon$$

share|cite|improve this answer

answered Jan 5 '18 at 12:58

mechanodroidmechanodroid

28.9k62648

$endgroup$

share|cite|improve this answer

answered Jan 5 '18 at 12:58

mechanodroidmechanodroid

28.9k62648

answered Jan 5 '18 at 12:58

mechanodroidmechanodroid

28.9k62648

answered Jan 5 '18 at 12:58

mechanodroidmechanodroid

28.9k62648

0

$begingroup$

For $xin [1,3]quad |(x-2)(x^2+2x+4)|le 19|x-2|<epsilon$

share|cite|improve this answer

answered Jan 5 '18 at 13:05

StuStu

1,1951414

$endgroup$

add a comment | 

0

$begingroup$

For $xin [1,3]quad |(x-2)(x^2+2x+4)|le 19|x-2|<epsilon$

share|cite|improve this answer

answered Jan 5 '18 at 13:05

StuStu

1,1951414

$endgroup$

add a comment | 

$begingroup$

For $xin [1,3]quad |(x-2)(x^2+2x+4)|le 19|x-2|<epsilon$

share|cite|improve this answer

answered Jan 5 '18 at 13:05

StuStu

1,1951414

$endgroup$

share|cite|improve this answer

answered Jan 5 '18 at 13:05

StuStu

1,1951414

answered Jan 5 '18 at 13:05

StuStu

1,1951414

answered Jan 5 '18 at 13:05

StuStu

1,1951414