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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!

S4M1R
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2 Answers2

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$$ \begin{split} \mathop {\lim }\limits_{x \to - 3} \frac{{4x + 12}} {{3{x^3} - 27x}} &= \mathop {\lim }\limits_{x \to - 3} \frac{{4\left( {x + 3} \right)}} {{3x\left( {{x^2} - 9} \right)}}\\ &= \mathop {\lim }\limits_{x \to - 3} \frac{{4\left( {x + 3} \right)}} {{3x\left( {x - 3} \right)\left( {x + 3} \right)}} \\ &= \mathop {\lim }\limits_{x \to - 3} \frac{4} {{3x\left( {x - 3} \right)}} \\ &= \frac{4} {{3\cdot\left( { - 3} \right)\cdot\left( { - 3 - 3} \right)}} = \frac{2} {{27}} \end{split} $$

Daniele Tampieri
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Baily
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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.

Andrei Rykhalski
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