Single Variable Calculus, Chapter 2, 2.3, Section 2.3, Problem 27

Determine the $\displaystyle \lim \limits_{x \to 16} \frac{4 - \sqrt{x}}{16x - x^2}$, if it exists.


$
\begin{equation}
\begin{aligned}

& \lim \limits_{x \to 16} \frac{4 - \sqrt{x}}{16x - x^2} \cdot \frac{4 + \sqrt{x}}{4 + \sqrt{x}}
= \lim \limits_{x \to 16} \frac{16 - x}{(16x - x^2)(4 + \sqrt{x})}
&& \text{ Multiply the numerator and the denominator by $4 + \sqrt{x}$.}\\
\\
& \lim \limits_{x \to 16} \frac{16 - x}{(16x - x^2)(4 + \sqrt{x})}
= \lim \limits_{x \to 16} \frac{\cancel{16 - x}}{(x)\cancel{(16 - x)} (4 + \sqrt{x})}
&& \text{Get the factor and cancel out like terms.}\\
\\
& \lim \limits_{x \to 16} \frac{1}{x(4 + \sqrt{x})}
= \frac{1 }{16 (4 + \sqrt{16})} = \frac{1}{16(4+4)} = \frac{1}{16(8)} = \frac{1}{128}
&& \text{ Substitute value of $x$ and simplify}\\
\\
\\
& \fbox{$ \lim \limits_{x \to 16} \displaystyle \frac{4 - \sqrt{x}}{16x - x^2} = \frac{1}{128}$}


\end{aligned}
\end{equation}
$