Question

Evalauate the limit, if it exists.

\(\mathop {{\bf{lim}}}\limits_{u \to - {\bf{1}}} \frac{{u + {\bf{1}}}}{{{u^{\bf{3}}} + {\bf{1}}}}\)

Short answer

The value of the limit is \(\frac{1}{3}\).

Step-by-step solution

Simplify the limit by using the factorization

Factorize the denominator of the expression\(\mathop {\lim }\limits_{u \to - 1} \frac{{u + 1}}{{{u^3} + 1}}\).

\(\begin{aligned}\mathop {\lim }\limits_{u \to - 1} \frac{{u + 1}}{{{u^3} + 1}} &= \mathop {\lim }\limits_{u \to - 1} \frac{{u + 1}}{{\left( {u + 1} \right)\left( {{u^2} - u + 1} \right)}}\\ &= \mathop {\lim }\limits_{u \to - 1} \frac{1}{{{u^2} - u + 1}}\end{aligned}\)

Evaluate the limit

For the expression \(\mathop {\lim }\limits_{u \to - 1} \frac{1}{{{u^2} - u + 1}}\), Quotient law is applicable.

According to the Quotient law, \(\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{f\left( x \right)}} = \frac{{\mathop {\lim }\limits_{x \to a} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to a} g\left( x \right)}}\), where \(\mathop {\lim }\limits_{x \to a} f\left( x \right)\) and \(\mathop {\lim }\limits_{x \to a} g\left( x \right)\) exists and \(\mathop {\lim }\limits_{x \to a} g\left( x \right) \ne 0\).

Solve the expression \(\mathop {\lim }\limits_{u \to - 1} \frac{1}{{{u^2} - u + 1}}\).

\(\begin{aligned}\mathop {\lim }\limits_{u \to - 1} \frac{1}{{{u^2} - u + 1}} &= \frac{{\mathop {\lim }\limits_{u \to - 1} \left( 1 \right)}}{{\mathop {\lim }\limits_{u \to - 1} \left( {{u^2} - u + 1} \right)}}\\ &= \frac{1}{{{{\left( { - 1} \right)}^2} - \left( { - 1} \right) + 1}}\\ &= \frac{1}{3}\end{aligned}\)

Thus, the value of the limit is \(\frac{1}{3}\).