Question

Evalauate the limit, if it exists.

\(\mathop {{\bf{lim}}}\limits_{t \to {\bf{3}}} \frac{{{t^{\bf{3}}} - {\bf{27}}}}{{{t^{\bf{2}}} - {\bf{9}}}}\)

Short answer

The value of the limit is \(\frac{9}{2}\).

Step-by-step solution

Simplify the limit by using the factorization

Factorizethe numerator and denominator of the expression \(\mathop {\lim }\limits_{t \to 3} \frac{{{t^3} - 27}}{{{t^2} - 9}}\).

\(\begin{aligned}\mathop {\lim }\limits_{t \to 3} \frac{{{t^3} - 27}}{{{t^2} - 9}} &= \mathop {\lim }\limits_{t \to 3} \frac{{\left( {t - 3} \right)\left( {{t^2} + 3t + 9} \right)}}{{\left( {t - 3} \right)\left( {t + 3} \right)}}\\ &= \mathop {\lim }\limits_{t \to 3} \frac{{{t^2} + 3t + 9}}{{t + 3}}\end{aligned}\)

Evaluate the limit

For the expression \(\mathop {\lim }\limits_{t \to 3} \frac{{{t^2} + 3t + 9}}{{t + 3}}\), 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 \(\mathop {\lim }\limits_{t \to 3} \frac{{{t^2} + 3t + 9}}{{t + 3}}\).

\(\begin{aligned}\mathop {\lim }\limits_{t \to 3} \frac{{{t^2} + 3t + 9}}{{t + 3}} &= \frac{{\mathop {\lim }\limits_{t \to 3} \left( {{t^2} + 3t + 9} \right)}}{{\mathop {\lim }\limits_{t \to 3} \left( {t + 3} \right)}}\\ &= \frac{{\mathop {\lim }\limits_{t \to 3} \left( {{t^2}} \right) + \mathop {\lim }\limits_{t \to 3} \left( {3t} \right) + \mathop {\lim }\limits_{t \to 3} \left( 9 \right)}}{{\mathop {\lim }\limits_{t \to 3} \left( t \right) + \mathop {\lim }\limits_{t \to 3} \left( 3 \right)}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{\rm{Sum}}\,\,{\rm{law}}} \right)\\ &= \frac{{{3^2} + 3\left( 3 \right) + 9}}{{3 + 3}}\,\\ &= \frac{{27}}{6}\\ &= \frac{9}{2}\end{aligned}\)

Thus, the value of the limit is \(\frac{9}{2}\).