## Precalculus (6th Edition) Blitzer

The limit of a product, $\left( f\left( x \right)\cdot g\left( x \right) \right)$ is $\underset{x\to a}{\mathop{\lim }}\,\left( f\left( x \right)\cdot g\left( x \right) \right)=\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)\cdot \underset{x\to a}{\mathop{\lim }}\,g\left( x \right)$.
For finding the limit of a product, first find the limit of each function in the product. Then multiply each of these limits. In other words, the limit of the product of two functions equals the product of their limits. In limit notation, If $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=L$ and $\underset{x\to a}{\mathop{\lim }}\,g\left( x \right)=M$, then \begin{align} & \underset{x\to a}{\mathop{\lim }}\,\left( f\left( x \right)\cdot g\left( x \right) \right)=\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)\cdot \underset{x\to a}{\mathop{\lim }}\,g\left( x \right) \\ & =L\cdot M \end{align} For example: Let $f\left( x \right)=x$ and $g\left( x \right)=2$. To find the limit of the product of $f\left( x \right)$ and $g\left( x \right)$, $\underset{x\to 2}{\mathop{\lim }}\,\left( f\left( x \right)\cdot g\left( x \right) \right)$, \begin{align} & \underset{x\to 2}{\mathop{\lim }}\,\left( f\left( x \right)\cdot g\left( x \right) \right)=\underset{x\to 2}{\mathop{\lim }}\,f\left( x \right)\cdot \underset{x\to 2}{\mathop{\lim }}\,g\left( x \right) \\ & =\underset{x\to 2}{\mathop{\lim }}\,x\cdot \underset{x\to 2}{\mathop{\lim }}\,2 \\ & =2\cdot 2 \\ & =4 \end{align}