evaluate the limit, if it exists. (if an answer does not exist, enter dne.)\n lim_{h\rightarrow0}\frac{sqrt{1…

Explanation:

Step1: Rationalize the numerator

Multiply the fraction by $\frac{\sqrt{16 + h}+4}{\sqrt{16 + h}+4}$. [ \begin{align*} &\lim_{h\rightarrow0}\frac{\sqrt{16 + h}-4}{h}\times\frac{\sqrt{16 + h}+4}{\sqrt{16 + h}+4}\ =&\lim_{h\rightarrow0}\frac{(\sqrt{16 + h})^2-4^2}{h(\sqrt{16 + h}+4)}\ =&\lim_{h\rightarrow0}\frac{16 + h - 16}{h(\sqrt{16 + h}+4)}\ =&\lim_{h\rightarrow0}\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\rightarrow0}\frac{h}{h(\sqrt{16 + h}+4)}=\lim_{h\rightarrow0}\frac{1}{\sqrt{16 + h}+4} ]

Step3: Evaluate the limit

Substitute $h = 0$ into the simplified function. [ \lim_{h\rightarrow0}\frac{1}{\sqrt{16 + h}+4}=\frac{1}{\sqrt{16+0}+4}=\frac{1}{4 + 4}=\frac{1}{8} ]

Answer:

$\frac{1}{8}$