Mixing
How to I solve this limit to -3
$begingroup$
I need help solving this limit $lim limits_{x to -3} frac{4x+12}{3x^3-27x}$.
I know that I am suppose to factor the function and then plug in -3 to calculate the result.
$lim limits_{x to -3} frac{4(x+3)}{3x^3-27x}$. But don't know how to factor $3x^3-27x$.
How do I factor $3x^3-27x$? I'm I on the right path?
Thanks!
limits
asked Nov 6 '14 at 10:41
S4M1RS4M1R
394127
$endgroup$
add a comment |
$begingroup$
I need help solving this limit $lim limits_{x to -3} frac{4x+12}{3x^3-27x}$.
I know that I am suppose to factor the function and then plug in -3 to calculate the result.
$lim limits_{x to -3} frac{4(x+3)}{3x^3-27x}$. But don't know how to factor $3x^3-27x$.
How do I factor $3x^3-27x$? I'm I on the right path?
Thanks!
limits
asked Nov 6 '14 at 10:41
S4M1RS4M1R
394127
$endgroup$
$begingroup$
$3x^3-27x=3x(x^2-9)=3x(x+3)(x-3)$ could help
$endgroup$
– Claude Leibovici
Nov 6 '14 at 10:47
add a comment |
$begingroup$
I need help solving this limit $lim limits_{x to -3} frac{4x+12}{3x^3-27x}$.
I know that I am suppose to factor the function and then plug in -3 to calculate the result.
$lim limits_{x to -3} frac{4(x+3)}{3x^3-27x}$. But don't know how to factor $3x^3-27x$.
How do I factor $3x^3-27x$? I'm I on the right path?
Thanks!
limits
asked Nov 6 '14 at 10:41
S4M1RS4M1R
394127
$endgroup$
I need help solving this limit $lim limits_{x to -3} frac{4x+12}{3x^3-27x}$.
I know that I am suppose to factor the function and then plug in -3 to calculate the result.
$lim limits_{x to -3} frac{4(x+3)}{3x^3-27x}$. But don't know how to factor $3x^3-27x$.
How do I factor $3x^3-27x$? I'm I on the right path?
Thanks!
limits
limits
asked Nov 6 '14 at 10:41
S4M1RS4M1R
394127
asked Nov 6 '14 at 10:41
S4M1RS4M1R
394127
asked Nov 6 '14 at 10:41
S4M1RS4M1R
394127
asked Nov 6 '14 at 10:41
S4M1RS4M1R
394127
asked Nov 6 '14 at 10:41
S4M1RS4M1R
394127
394127
$begingroup$
$3x^3-27x=3x(x^2-9)=3x(x+3)(x-3)$ could help
$endgroup$
– Claude Leibovici
Nov 6 '14 at 10:47
add a comment |
$begingroup$
$3x^3-27x=3x(x^2-9)=3x(x+3)(x-3)$ could help
$endgroup$
– Claude Leibovici
Nov 6 '14 at 10:47
$begingroup$
$3x^3-27x=3x(x^2-9)=3x(x+3)(x-3)$ could help
$endgroup$
– Claude Leibovici
Nov 6 '14 at 10:47
$begingroup$
$3x^3-27x=3x(x^2-9)=3x(x+3)(x-3)$ could help
$endgroup$
– Claude Leibovici
Nov 6 '14 at 10:47
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$$
begin{split}
mathop {lim }limits_{x to - 3} frac{{4x + 12}}
{{3{x^3} - 27x}} &= mathop {lim }limits_{x to - 3} frac{{4left( {x + 3} right)}}
{{3xleft( {{x^2} - 9} right)}}\
&= mathop {lim }limits_{x to - 3} frac{{4left( {x + 3} right)}}
{{3xleft( {x - 3} right)left( {x + 3} right)}} \
&= mathop {lim }limits_{x to - 3} frac{4}
{{3xleft( {x - 3} right)}} \
&= frac{4}
{{3cdotleft( { - 3} right)cdotleft( { - 3 - 3} right)}} = frac{2}
{{27}}
end{split}
$$
edited Jan 20 at 17:14
Daniele Tampieri
2,3372922
answered Nov 6 '14 at 10:47
User3101User3101
1,301510
$endgroup$
add a comment |
$begingroup$
As your limit is of type $[frac 00]$, you may use l'Hospitale rule, it's the most simple way to find this limit. You should just differentiate numerator and denominator and the limit will be the same. Hope you can continue by yourself from this.
answered Nov 6 '14 at 10:46
Andrei RykhalskiAndrei Rykhalski
1,237614
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
oldest
votes
$begingroup$
$$
begin{split}
mathop {lim }limits_{x to - 3} frac{{4x + 12}}
{{3{x^3} - 27x}} &= mathop {lim }limits_{x to - 3} frac{{4left( {x + 3} right)}}
{{3xleft( {{x^2} - 9} right)}}\
&= mathop {lim }limits_{x to - 3} frac{{4left( {x + 3} right)}}
{{3xleft( {x - 3} right)left( {x + 3} right)}} \
&= mathop {lim }limits_{x to - 3} frac{4}
{{3xleft( {x - 3} right)}} \
&= frac{4}
{{3cdotleft( { - 3} right)cdotleft( { - 3 - 3} right)}} = frac{2}
{{27}}
end{split}
$$
edited Jan 20 at 17:14
Daniele Tampieri
2,3372922
answered Nov 6 '14 at 10:47
User3101User3101
1,301510
$endgroup$
add a comment |
$begingroup$
$$
begin{split}
mathop {lim }limits_{x to - 3} frac{{4x + 12}}
{{3{x^3} - 27x}} &= mathop {lim }limits_{x to - 3} frac{{4left( {x + 3} right)}}
{{3xleft( {{x^2} - 9} right)}}\
&= mathop {lim }limits_{x to - 3} frac{{4left( {x + 3} right)}}
{{3xleft( {x - 3} right)left( {x + 3} right)}} \
&= mathop {lim }limits_{x to - 3} frac{4}
{{3xleft( {x - 3} right)}} \
&= frac{4}
{{3cdotleft( { - 3} right)cdotleft( { - 3 - 3} right)}} = frac{2}
{{27}}
end{split}
$$
edited Jan 20 at 17:14
Daniele Tampieri
2,3372922
answered Nov 6 '14 at 10:47
User3101User3101
1,301510
$endgroup$
add a comment |
$begingroup$
$$
begin{split}
mathop {lim }limits_{x to - 3} frac{{4x + 12}}
{{3{x^3} - 27x}} &= mathop {lim }limits_{x to - 3} frac{{4left( {x + 3} right)}}
{{3xleft( {{x^2} - 9} right)}}\
&= mathop {lim }limits_{x to - 3} frac{{4left( {x + 3} right)}}
{{3xleft( {x - 3} right)left( {x + 3} right)}} \
&= mathop {lim }limits_{x to - 3} frac{4}
{{3xleft( {x - 3} right)}} \
&= frac{4}
{{3cdotleft( { - 3} right)cdotleft( { - 3 - 3} right)}} = frac{2}
{{27}}
end{split}
$$
edited Jan 20 at 17:14
Daniele Tampieri
2,3372922
answered Nov 6 '14 at 10:47
User3101User3101
1,301510
$endgroup$
$$
begin{split}
mathop {lim }limits_{x to - 3} frac{{4x + 12}}
{{3{x^3} - 27x}} &= mathop {lim }limits_{x to - 3} frac{{4left( {x + 3} right)}}
{{3xleft( {{x^2} - 9} right)}}\
&= mathop {lim }limits_{x to - 3} frac{{4left( {x + 3} right)}}
{{3xleft( {x - 3} right)left( {x + 3} right)}} \
&= mathop {lim }limits_{x to - 3} frac{4}
{{3xleft( {x - 3} right)}} \
&= frac{4}
{{3cdotleft( { - 3} right)cdotleft( { - 3 - 3} right)}} = frac{2}
{{27}}
end{split}
$$
edited Jan 20 at 17:14
Daniele Tampieri
2,3372922
answered Nov 6 '14 at 10:47
User3101User3101
1,301510
edited Jan 20 at 17:14
Daniele Tampieri
2,3372922
edited Jan 20 at 17:14
Daniele Tampieri
2,3372922
edited Jan 20 at 17:14
Daniele Tampieri
2,3372922
2,3372922
answered Nov 6 '14 at 10:47
User3101User3101
1,301510
answered Nov 6 '14 at 10:47
User3101User3101
1,301510
answered Nov 6 '14 at 10:47
User3101User3101
1,301510
1,301510
add a comment |
add a comment |
$begingroup$
As your limit is of type $[frac 00]$, you may use l'Hospitale rule, it's the most simple way to find this limit. You should just differentiate numerator and denominator and the limit will be the same. Hope you can continue by yourself from this.
answered Nov 6 '14 at 10:46
Andrei RykhalskiAndrei Rykhalski
1,237614
$endgroup$
add a comment |
$begingroup$
As your limit is of type $[frac 00]$, you may use l'Hospitale rule, it's the most simple way to find this limit. You should just differentiate numerator and denominator and the limit will be the same. Hope you can continue by yourself from this.
answered Nov 6 '14 at 10:46
Andrei RykhalskiAndrei Rykhalski
1,237614
$endgroup$
add a comment |
$begingroup$
As your limit is of type $[frac 00]$, you may use l'Hospitale rule, it's the most simple way to find this limit. You should just differentiate numerator and denominator and the limit will be the same. Hope you can continue by yourself from this.
answered Nov 6 '14 at 10:46
Andrei RykhalskiAndrei Rykhalski
1,237614
$endgroup$
As your limit is of type $[frac 00]$, you may use l'Hospitale rule, it's the most simple way to find this limit. You should just differentiate numerator and denominator and the limit will be the same. Hope you can continue by yourself from this.
answered Nov 6 '14 at 10:46
Andrei RykhalskiAndrei Rykhalski
1,237614
answered Nov 6 '14 at 10:46
Andrei RykhalskiAndrei Rykhalski
1,237614
answered Nov 6 '14 at 10:46
Andrei RykhalskiAndrei Rykhalski
1,237614
answered Nov 6 '14 at 10:46
Andrei RykhalskiAndrei Rykhalski
1,237614
1,237614
add a comment |
add a comment |
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$begingroup$
$3x^3-27x=3x(x^2-9)=3x(x+3)(x-3)$ could help
$endgroup$
– Claude Leibovici
Nov 6 '14 at 10:47