## Precalculus (6th Edition) Blitzer

The limit of a difference, $\left( f\left( x \right)-g\left( x \right) \right)$ is $\underset{x\to a}{\mathop{\lim }}\,\left( f\left( x \right)-g\left( x \right) \right)=\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)-\underset{x\to a}{\mathop{\lim }}\,g\left( x \right)$.
For finding the limit of a difference, first find the limit of each function in the difference. Then subtract each of these limits. In other words, the limit of the difference of two functions equals the difference 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)-g\left( x \right) \right)=\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)-\underset{x\to a}{\mathop{\lim }}\,g\left( x \right) \\ & =L-M \end{align} For example: Let $f\left( x \right)=x$ and $g\left( x \right)=2$. To find the limit of the difference of $f\left( x \right)$ and $g\left( x \right)$, $\underset{x\to 2}{\mathop{\lim }}\,\left( f\left( x \right)-g\left( x \right) \right)$, \begin{align} & \underset{x\to 2}{\mathop{\lim }}\,\left( f\left( x \right)-g\left( x \right) \right)=\underset{x\to 2}{\mathop{\lim }}\,f\left( x \right)-\underset{x\to 2}{\mathop{\lim }}\,g\left( x \right) \\ & =\underset{x\to 2}{\mathop{\lim }}\,x-\underset{x\to 2}{\mathop{\lim }}\,2 \\ & =2-2 \\ & =0 \end{align}