Question

Evalauate the limit, if it exists.

\(\mathop {{\bf{lim}}}\limits_{x \to - {\bf{2}}} \frac{{{x^{\bf{2}}} - x - {\bf{6}}}}{{{\bf{3}}{x^{\bf{2}}} + {\bf{5}}x - {\bf{2}}}}\)

Short answer

The value of the limt is \(\frac{5}{7}\).

Step-by-step solution

Simplify the limit by using the factorization

Factorizethe numerator and denominator of the expression\(\mathop {\lim }\limits_{x \to - 2} \frac{{{x^2} - x - 6}}{{3{x^2} + 5x - 2}}\).

\(\begin{aligned}\mathop {\lim }\limits_{x \to - 2} \frac{{{x^2} - x - 6}}{{3{x^2} + 5x - 2}} &= \mathop {\lim }\limits_{x \to - 2} \frac{{\left( {x - 3} \right)\left( {x + 2} \right)}}{{\left( {3x - 1} \right)\left( {x + 2} \right)}}\\ &= \mathop {\lim }\limits_{x \to - 2} \frac{{x - 3}}{{3x - 1}}\end{aligned}\)

Evaluate the limit

Forthe expression \(\mathop {\lim }\limits_{x \to - 2} \frac{{x - 3}}{{3x - 1}}\), 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 \(\mathop {\lim }\limits_{x \to - 2} \frac{{x - 3}}{{3x - 1}}\).

\(\begin{aligned}\mathop {\lim }\limits_{x \to - 2} \frac{{x - 3}}{{3x - 1}} &= \frac{{\mathop {\lim }\limits_{x \to - 2} \left( {x - 3} \right)}}{{\mathop {\lim }\limits_{x \to - 2} \left( {3x - 1} \right)}}\\ &= \frac{{\mathop {\lim }\limits_{x \to - 2} \left( x \right) - \mathop {\lim }\limits_{x \to - 2} \left( 3 \right)}}{{\mathop {\lim }\limits_{x \to - 2} \left( {3x} \right) - \mathop {\lim }\limits_{x \to - 2} \left( 1 \right)}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\rm{Difference}}\;{\rm{law}}} \right)\\ &= \frac{{ - 2 - 3}}{{3\left( { - 2} \right) - 1}}\,\\ &= \frac{{ - 5}}{{ - 7}}\\ &= \frac{5}{7}\end{aligned}\)

Thus, the value of the limit is\(\frac{5}{7}\).