# How do you find the limit of f(x)= (3x^2 - 9x - 12) / (x^2 + 2x -24) as x approaches 4?

Apr 21, 2015

Use l'Hospital's rule.

When both the numerator and the denominator give 0 when the limiting x-value is plugged into them, you should you l'Hospital's rule to find the answer. This is how you do it:

1. Take the derivative of the top and bottom.
2. The original limit is equal to the ratio of the two functions you have just obtained by taking the derivatives.
Apr 21, 2015

$2.1$

$\textcolor{b l u e}{S o l u t i o n}$

$\frac{3 {x}^{2} - 9 x - 12}{{x}^{2} + 2 x - 24}$

$\implies \frac{3 {x}^{2} - 12 x + 3 x - 12}{{x}^{2} + 6 x - 4 x - 24}$

$\implies \frac{3 x \left(x - 4\right) + 3 \left(x - 4\right)}{x \left(x + 6\right) - 4 \left(x + 6\right)}$

$\implies \frac{\left(3 x + 3\right) \left(x - 4\right)}{\left(x + 6\right) \left(x - 4\right)}$

$\implies \frac{\left(3 x + 3\right) \left(\cancel{x - 4}\right)}{\left(x + 6\right) \left(\cancel{x - 4}\right)}$

$\implies \frac{3 \left(x + 3\right)}{x + 6}$

Now, applying limit as $x \to 4$

$\implies \frac{{\lim}_{x \to 4} \left(3 \left(x + 3\right)\right)}{{\lim}_{x \to 4} \left(x + 6\right)}$

$\implies \frac{3 {\lim}_{x \to 4} \left(x + 3\right)}{{\lim}_{x \to 4} \left(x + 6\right)}$

$\implies \frac{3 \left(4 + 3\right)}{4 + 6}$

$\implies \frac{3 \left(7\right)}{10}$

$\implies \frac{21}{10}$

$\implies 2.1$