Precalculus (6th Edition) Blitzer

The limit $\underset{x\to 4}{\mathop{\lim }}\,\left( \frac{1}{x}-\frac{1}{4} \right)\left( \frac{1}{x-4} \right)$ is equal to $\frac{-1}{16}$.
Consider the provided limit, $\underset{x\to 4}{\mathop{\lim }}\,\left( \frac{1}{x}-\frac{1}{4} \right)\left( \frac{1}{x-4} \right)$ Simplify it as follows: Take the L.C.M of 4 and $x$, which is $4x$. \begin{align} & \underset{x\to 4}{\mathop{\lim }}\,\left( \frac{1}{x}-\frac{1}{4} \right)\left( \frac{1}{x-4} \right)=\underset{x\to 4}{\mathop{\lim }}\,\left( \frac{4-x}{4x} \right)\left( \frac{1}{x-4} \right) \\ & =\underset{x\to 4}{\mathop{\lim }}\,\left( \frac{-\left( x-4 \right)}{4x} \right)\left( \frac{1}{x-4} \right) \\ & =\underset{x\to 4}{\mathop{\lim }}\,\frac{\left( -1 \right)}{4x} \end{align} Use quotient property of limits: \begin{align} & \underset{x\to 4}{\mathop{\lim }}\,\left( \frac{1}{x}-\frac{1}{4} \right)\left( \frac{1}{x-4} \right)=\underset{x\to 4}{\mathop{\lim }}\,\frac{\left( -1 \right)}{4x} \\ & =\frac{\underset{x\to 4}{\mathop{\lim }}\,\left( -1 \right)}{\underset{x\to 4}{\mathop{\lim }}\,4x} \end{align} Using limit property $\underset{x\to a}{\mathop{\lim }}\,c=c\text{, where }c\text{ is constant}$ \begin{align} & \underset{x\to 4}{\mathop{\lim }}\,\left( \frac{1}{x}-\frac{1}{4} \right)\left( \frac{1}{x-4} \right)=\underset{x\to 4}{\mathop{\lim }}\,\frac{\left( -1 \right)}{4x} \\ & =\frac{\underset{x\to 4}{\mathop{\lim }}\,\left( -1 \right)}{\underset{x\to 4}{\mathop{\lim }}\,4x} \\ & =\frac{-1}{4\times 4} \\ & =\frac{-1}{16} \end{align} Thus, $\underset{x\to 4}{\mathop{\lim }}\,\left( \frac{1}{x}-\frac{1}{4} \right)\left( \frac{1}{x-4} \right)=\frac{-1}{16}$