Question

evaluate the limit, if it exists. (if an answer does not exist, enter dne.)
lim_{h
ightarrow0}\frac{sqrt{16 + h}-4}{h}

Explanation:

Step1: Rationalize the numerator

Multiply the fraction by $\frac{\sqrt{16 + h}+4}{\sqrt{16 + h}+4}$.
\[

$$\begin{align*} &\lim_{h ightarrow0}\frac{\sqrt{16 + h}-4}{h}\times\frac{\sqrt{16 + h}+4}{\sqrt{16 + h}+4}\\ =&\lim_{h ightarrow0}\frac{(\sqrt{16 + h})^2-4^2}{h(\sqrt{16 + h}+4)}\\ =&\lim_{h ightarrow0}\frac{16 + h - 16}{h(\sqrt{16 + h}+4)}\\ =&\lim_{h ightarrow0}\frac{h}{h(\sqrt{16 + h}+4)} \end{align*}$$

\]

Step2: Simplify the fraction

Cancel out the common factor $h$ in the numerator and denominator.
\[
\lim_{h
ightarrow0}\frac{h}{h(\sqrt{16 + h}+4)}=\lim_{h
ightarrow0}\frac{1}{\sqrt{16 + h}+4}
\]

Step3: Evaluate the limit

Substitute $h = 0$ into the simplified function.
\[
\lim_{h
ightarrow0}\frac{1}{\sqrt{16 + h}+4}=\frac{1}{\sqrt{16+0}+4}=\frac{1}{4 + 4}=\frac{1}{8}
\]

Answer:

$\frac{1}{8}$