Question

Evaluate the limit, if it exists.

\(\mathop {{\bf{lim}}}\limits_{x \to - {\bf{3}}} \frac{{{x^{\bf{2}}} + {\bf{3}}x}}{{{x^{\bf{2}}} - x - {\bf{12}}}}\)

Short answer

The value of the limit is \(\frac{3}{7}\).

Step-by-step solution

Step 1:Simplify the limit by using the factorization

Factorize the numerator and denominator of the expression \(\mathop {\lim }\limits_{x \to - 3} \frac{{{x^2} + 3x}}{{{x^2} - x - 12}}\).

\(\begin{aligned}\mathop {\lim }\limits_{x \to - 3} \frac{{{x^2} + 3x}}{{{x^2} - x - 12}} &= \mathop {\lim }\limits_{x \to - 3} \frac{{x\left( {x + 3} \right)}}{{\left( {x - 4} \right)\left( {x + 3} \right)}}\\ &= \mathop {\lim }\limits_{x \to - 3} \frac{x}{{x - 4}}\end{aligned}\)

Evaluate the limit

For the expression \(\mathop {\lim }\limits_{x \to - 3} \frac{x}{{x - 4}}\)quotient lawis 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 \(\mathop {\lim }\limits_{x \to - 3} \frac{x}{{x - 4}}\).

\(\begin{aligned}\mathop {\lim }\limits_{x \to - 3} \frac{x}{{x - 4}} &= \frac{{\mathop {\lim }\limits_{x \to - 3} \left( x \right)}}{{\mathop {\lim }\limits_{x \to - 3} \left( {x - 4} \right)}}\\ &= \frac{{ - 3}}{{ - 3 - 4}}\,\\ &= \frac{3}{7}\end{aligned}\)

Thus, the value of the limit is \(\frac{3}{7}\).