# How do you find the limit of (3x^2-x-10)/(x^2+5x-14) as x approaches 2?

May 30, 2018

$\frac{11}{9}$

#### Explanation:

Observe that
$3 {x}^{2} - x - 10 = \left(3 x + 5\right) \left(x - 2\right)$
${x}^{2} + 5 x - 14 = \left(x + 7\right) \left(x - 2\right)$
So we get

${\lim}_{x \to 2} \frac{3 {x}^{2} - x - 10}{{x}^{2} + 5 x - 1} = {\lim}_{x \to 2} \frac{3 x + 5}{x + 7} = \frac{11}{9}$

May 30, 2018

${\lim}_{x \to 2} \frac{3 {x}^{2} - x - 10}{{x}^{2} + 5 x - 14} = \frac{11}{9}$

#### Explanation:

Method 1): factorize the polynomials

As both numerator and denominator vanish for $x = 2$ they can be divided by $\left(x - 2\right)$:

$3 {x}^{2} - x - 10 = \left(3 x + 5\right) \left(x - 2\right)$

${x}^{2} + 5 x - 14 = \left(x + 7\right) \left(x - 2\right)$

Then:

${\lim}_{x \to 2} \frac{3 {x}^{2} - x - 10}{{x}^{2} + 5 x - 14} = {\lim}_{x \to 2} \frac{\left(3 x + 5\right) \left(x - 2\right)}{\left(x + 7\right) \left(x - 2\right)}$

${\lim}_{x \to 2} \frac{3 {x}^{2} - x - 10}{{x}^{2} + 5 x - 14} = {\lim}_{x \to 2} \frac{3 x + 5}{x + 7}$

${\lim}_{x \to 2} \frac{3 {x}^{2} - x - 10}{{x}^{2} + 5 x - 14} = \frac{11}{9}$

Method 2): L'Hospital's rule:

As both numerator and denominator vanish for $x = 2$:

${\lim}_{x \to 2} \frac{3 {x}^{2} - x - 10}{{x}^{2} + 5 x - 14} = {\lim}_{x \to 2} \frac{\frac{d}{\mathrm{dx}} \left(3 {x}^{2} - x - 10\right)}{\frac{d}{\mathrm{dx}} \left({x}^{2} + 5 x - 14\right)}$

${\lim}_{x \to 2} \frac{3 {x}^{2} - x - 10}{{x}^{2} + 5 x - 14} = {\lim}_{x \to 2} \frac{6 x - 1}{2 x + 5} = \frac{11}{9}$