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}$} \end{aligned} \end{equation}$