Answer
$\frac{1}{3}$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{u \to - 1} \frac{{u + 1}}{{{u^3} + 1}} \cr
& {\text{Evaluate the limit by using the Direct Substitution Property}} \cr
& \mathop {\lim }\limits_{u \to a} f\left( u \right) = f\left( a \right) \cr
& \cr
& {\text{Therefore}}{\text{,}} \cr
& \mathop {\lim }\limits_{u \to - 1} \frac{{u + 1}}{{{u^3} + 1}} = \frac{{\left( { - 1} \right) + 1}}{{{{\left( { - 1} \right)}^3} + 1}} \cr
& \cr
& {\text{Simplifying}} \cr
& \mathop {\lim }\limits_{u \to - 1} \frac{{u + 1}}{{{u^3} + 1}} = \frac{{\left( { - 1} \right) + 1}}{{{{\left( { - 1} \right)}^3} + 1}} = \frac{0}{0}{\text{ Indeterminate}} \cr
& \cr
& {\text{Factoring the denominator}} \cr
& \mathop {\lim }\limits_{u \to - 1} \frac{{u + 1}}{{{u^3} + 1}} = \mathop {\lim }\limits_{u \to - 1} \frac{{u + 1}}{{\left( {u + 1} \right)\left( {{u^2} - u + 1} \right)}} \cr
& {\text{Cancel the common factor }}u + 1 \cr
& = \mathop {\lim }\limits_{u \to - 1} \frac{1}{{{u^2} - u + 1}} \cr
& \cr
& {\text{Evaluate the limit}} \cr
& \mathop {\lim }\limits_{u \to - 1} \frac{1}{{{u^2} - u + 1}} = \frac{1}{{{{\left( { - 1} \right)}^2} - \left( { - 1} \right) + 1}} = \frac{1}{3} \cr
& {\text{Then}} \cr
& \mathop {\lim }\limits_{u \to - 1} \frac{1}{{{u^2} - u + 1}} = \frac{1}{3} \cr} $$
