Question

Evaluate the limit and justify each step by indicating the appropriate Limit Law(s).

3. \(\mathop {lim}\limits_{v \to 2} \left( {{v^2} + 2v} \right)\left( {2{v^3} - 5} \right)\)

Short answer

The value of the limit is \(\mathop {lim}\limits_{v \to 2} \left( {{v^2} + 2v} \right)\left( {2{v^3} - 5} \right) = 88\).

Step-by-step solution

Apply Product laws

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

Apply the law in given function as:

\(\mathop {lim}\limits_{v \to 2} \left( {{v^2} + 2v} \right)\left( {2{v^3} - 5} \right) = \mathop {lim}\limits_{v \to 2} \left( {{v^2} + 2v} \right) \cdot \mathop {lim}\limits_{v \to 2} \left( {2{v^3} - 5} \right)\)

Apply Difference and sum laws

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

Apply the law as:

\(\mathop {lim}\limits_{v \to 2} \left( {{v^2} + 2v} \right) \cdot \mathop {lim}\limits_{v \to 2} \left( {2{v^3} - 5} \right) = \left( {\mathop {lim}\limits_{v \to 2} {v^2} + \mathop {lim}\limits_{v \to 2} 2v} \right) \cdot \left( {\mathop {lim}\limits_{v \to 2} 2{v^3} - \mathop {lim}\limits_{v \to 2} 5} \right)\)

Apply Constant Multiple laws

According the constant multiple law, \(\mathop {\lim }\limits_{x \to a} cf\left( x \right) = c\mathop {\lim }\limits_{x \to a} f\left( x \right)\), where \(\mathop {\lim }\limits_{x \to a} f\left( x \right)\) exists.

Apply the law as:

\(\left( {\mathop {lim}\limits_{v \to 2} {v^2} + \mathop {lim}\limits_{v \to 2} 2v} \right) \cdot \left( {\mathop {lim}\limits_{v \to 2} 2{v^3} - \mathop {lim}\limits_{v \to 2} 5} \right) = \left( {\mathop {lim}\limits_{v \to 2} {v^2} + 2\mathop {lim}\limits_{v \to 2} v} \right) \cdot \left( {2\mathop {lim}\limits_{v \to 2} {v^3} - \mathop {lim}\limits_{v \to 2} 5} \right)\)

Apply limit laws

According to the law,\(\mathop {\lim }\limits_{x \to a} x = a\),\(\mathop {\lim }\limits_{x \to a} x = a\) and\(\mathop {\lim }\limits_{x \to a} {x^n} = {a^n}\)where\(n\)is positive integer.

Apply the law, and the values as shown below:

\(\begin{array}{c}\left( {\mathop {lim}\limits_{v \to 2} {v^2} + 2\mathop {lim}\limits_{v \to 2} v} \right) \cdot \left( {2\mathop {lim}\limits_{v \to 2} {v^3} - \mathop {lim}\limits_{v \to 2} 5} \right) &=& \left( {{2^2} + 2\left( 2 \right)} \right) \cdot \left( {2 \cdot {{\left( 2 \right)}^3} - 5} \right)\\ &=& \left( {4 + 4} \right) \cdot \left( {16 - 5} \right)\\ &=& 8 \cdot 11\\ &=& 88\end{array}\)

Thus, the value of the limit is 88.