# How do you evaluate the limit (x^4+3x^3-x^2+x+4)/(x+1) as x approaches -1?

Oct 24, 2016

Factor and simplify the ratio.

#### Explanation:

Because $x = - 1$ makes the numerator $0$, we can be sure that $x - \left(- 1\right) = x + 1$ is a factor of the numerator.

Use division or trial and error to get

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

So,

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

$= {\lim}_{x \rightarrow - 1} \left({x}^{3} + 2 {x}^{2} - 3 x + 4\right) = - 1 + 2 + 3 + 4 = 8$

Oct 24, 2016

Use L'Hôpital's rule
${\lim}_{x \to - 1} \frac{{x}^{3} + 3 {x}^{3} - {x}^{2} + x + 4}{x + 1} = 8$

#### Explanation:

Because the expression becomes an indeterminate form, specifically $\frac{0}{0}$ when evaluated at the limit, L'Hôpital's rule applies. Please understand, though the implied division can be performed without a remainder, this is not a valid way to determine the limit.

The derivative of the denominator is 1, therefore, we do not need to write it into the resulting expression.

The derivative of the numerator is:

$4 {x}^{3} + 9 {x}^{2} - 2 x + 1$

${\lim}_{x \to - 1} 4 {x}^{3} + 9 {x}^{2} - 2 x + 1 = - 4 + 9 + 2 + 1 = 8$

L'Hôpital's rule stipulates that the limit of the original expression is the same.

${\lim}_{x \to - 1} \frac{{x}^{3} + 3 {x}^{3} - {x}^{2} + x + 4}{x + 1} = 8$