Answer
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 $\frac{-1}{16}$.
Work Step by Step
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, $\underset{x\to 4}{\mathop{\lim }}\,\left( \frac{1}{x}-\frac{1}{4} \right)\left( \frac{1}{x-4} \right)=\frac{-1}{16}$
