Differential and Integral Calculus by Feliciano and Uy, Exercise 1.3, Problem 6

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PROBLEM:

Evaluate $\displaystyle\lim\limits_{x\to 0}\:\frac{\sqrt{x+16}-4}{x}$ .

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SOLUTION:

A straight substitution of $x=0$ leads to the indeterminate form $\frac{0}{0}$ which is meaningless.

Therefore, to evaluate the limit of the given function, we proceed as follows.

\begin{align*} \\ \lim\limits_{x\to 0}\:\frac{\sqrt{x+16}-4}{x} & =\lim\limits_{x\to 0}\:\frac{\sqrt{x+16}-4}{x}\cdot \frac{\sqrt{x+16}+4}{\sqrt{x+16}+4}\\ \\ & =\lim\limits_{x\to 0}\:\frac{\left(x+16\right)-4^2}{x\left(\sqrt{x+16}+4\right)}\\ \\ & =\lim\limits_{x\to 0}\:\frac{x+16-16}{x\left(\sqrt{x+16}+4\right)}\\ \\ & =\lim\limits_{x\to 0}\:\frac{x}{x\left(\sqrt{x+16}+4\right)}\\ \\ & =\lim\limits_{x\to 0}\:\frac{1}{\sqrt{x+16}+4}\\ \\ & =\:\frac{1}{\sqrt{0+16}+4}\\ \\ & =\:\frac{1}{4+4}\\ \\ & =\:\frac{1}{8} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \\ \end{align*}

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