# How do you find the limit lim_(x->-4)(x^2+5x+4)/(x^2+3x-4) ?

Jul 29, 2014

$= \frac{3}{5}$

Explanation,

$= {\lim}_{x \to - 4} \frac{{x}^{2} + 5 x + 4}{{x}^{2} + 3 x - 4}$ , if we plug $x = - 4$, we get $\frac{0}{0}$ form

$= {\lim}_{x \to - 4} \frac{{x}^{2} + 4 x + x + 4}{{x}^{2} + 4 x - x - 4}$

$= {\lim}_{x \to - 4} \frac{x \left(x + 4\right) + 1 \left(x + 4\right)}{x \left(x + 4\right) - 1 \left(x + 4\right)}$

$= {\lim}_{x \to - 4} \frac{\left(x + 4\right) \left(x + 1\right)}{\left(x + 4\right) \left(x - 1\right)}$

$= {\lim}_{x \to - 4} \frac{\left(x + 1\right)}{\left(x - 1\right)}$

$= \frac{- 3}{-} 5$

$= \frac{3}{5}$