# How do you evaluate [(x^2) + 5x + 4] / [(x^2) + 3x -4] as x approaches -4?

Hence the limit is an undefined form $\frac{0}{0}$ we can apply L'Hopital law so we have that

lim_(x->-4) [x^2+5x+4]/[x^2+3x+4]= lim_(x->-4) (d[x^2+5x+4]/dx)/(d[x^2+3x+4]/dx)= lim_(x->-4) (2x+5)/(2x+3)=(2*(-4)+5)/(2*(-4)+3)= (-3)/(-5)=3/5

Mar 31, 2016

If you have not yet learned about derivatives and l'Hospital's rule, you can still find the limit.

#### Explanation:

${\lim}_{x \rightarrow - 4} \frac{{x}^{2} + 5 x + 4}{{x}^{2} + 3 x - 4}$ has indeterminate initial form. $- 4$ is a zero of both the numerator and denominator.

The numerator and denominator are polynomials, and we know from Algebra, that if $z$ is a zero of a polynomial, then $x - z$ is a factor of the polynomial.

So we know that $x - \left(- 4\right) = x + 4$ is a factor of both the numerator and the denominator of $f$.

So we can factor and reduce and try again to evaluate the limit.
Factoring is simplified by the fact that we already know one of the factors is $x + 4$

$\frac{{x}^{2} + 5 x + 4}{{x}^{2} + 3 x - 4} = \frac{\left(x + 4\right) \left(x + 1\right)}{\left(x + 4\right) \left(x - 1\right)}$.

Now for every $x$ other than $- 4$,

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

(for $x = - 4$ there is no number of the left, so there is nothing there to be equal or different from anything.)

The limit as $x$ approaches $- 4$ doesn't care what happens when $x$ actually equals $- 4$, so

${\lim}_{x \rightarrow - 4} \frac{{x}^{2} + 5 x + 4}{{x}^{2} + 3 x - 4} = {\lim}_{x \rightarrow - 4} \frac{\left(x + 4\right) \left(x + 1\right)}{\left(x + 4\right) \left(x - 1\right)}$

$= {\lim}_{x \rightarrow - 4} \frac{x + 1}{x - 1}$

We can evaluate this limit by substitution. We get

${\lim}_{x \rightarrow - 4} \frac{x + 1}{x - 1} = \frac{\left(- 4\right) + 1}{\left(- 4\right) - 1} = \frac{- 3}{- 5} = \frac{3}{5}$

So

${\lim}_{x \rightarrow - 4} \frac{{x}^{2} + 5 x + 4}{{x}^{2} + 3 x - 4} = \frac{3}{5}$