Question

Evalauate the limit, if it exists.

\(\mathop {{\bf{lim}}}\limits_{t \to {\bf{4}}} \frac{{{t^{\bf{2}}} - {\bf{2}}t - {\bf{8}}}}{{t - {\bf{4}}}}\)

Short answer

The value of the limit is 6.

Step-by-step solution

Step 1:Simplify the limit by using the factorization

Factorize the numerator of the expression\(\mathop {\lim }\limits_{t \to 4} \frac{{{t^2} - 2t - 8}}{{t - 4}}\).

\(\begin{aligned}\mathop {\lim }\limits_{t \to 4} \frac{{{t^2} - 2t - 8}}{{t - 4}} &= \mathop {\lim }\limits_{t \to 4} \frac{{\left( {t - 4} \right)\left( {t + 2} \right)}}{{t - 4}}\\ &= \mathop {\lim }\limits_{t \to 4} \left( {t + 2} \right)\end{aligned}\)

Evaluate the limit

For the limit\(\mathop {\lim }\limits_{t \to 4} \left( {t + 2} \right)\),sum law is applicable.

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

Solve the expression \(\mathop {\lim }\limits_{t \to 4} \left( {t + 2} \right)\).

\(\begin{aligned}\mathop {\lim }\limits_{t \to 4} \left( {t + 2} \right) &= \mathop {\lim }\limits_{t \to 4} \left( t \right) + \mathop {\lim }\limits_{t \to 4} 2\\ &= \left( {4 + 2} \right)\\ &= 6\end{aligned}\)

Thus, the value of the limit is 6.