Question

1. - / 3.84 points

evaluate the limit, if it exists. (if an answer does not exist, enter dne.)
\\(\lim_{h\to0}\frac{\sqrt{9 + h}-3}{h}\\)

Explanation:

Step1: Rationalize the numerator

Multiply the fraction by $\frac{\sqrt{9 + h}+3}{\sqrt{9 + h}+3}$.
\[

$$\begin{align*} &\lim_{h ightarrow0}\frac{\sqrt{9 + h}-3}{h}\times\frac{\sqrt{9 + h}+3}{\sqrt{9 + h}+3}\\ =&\lim_{h ightarrow0}\frac{(\sqrt{9 + h}-3)(\sqrt{9 + h}+3)}{h(\sqrt{9 + h}+3)} \end{align*}$$

\]
Using the difference - of - squares formula $(a - b)(a + b)=a^{2}-b^{2}$, we get $\lim_{h
ightarrow0}\frac{(9 + h)-9}{h(\sqrt{9 + h}+3)}$.

Step2: Simplify the numerator

Simplify the numerator $(9 + h)-9$ to $h$.
The limit becomes $\lim_{h
ightarrow0}\frac{h}{h(\sqrt{9 + h}+3)}$.

Step3: Cancel out the common factor

Cancel out the common factor $h$ in the numerator and denominator (since $h
eq0$ as we are taking the limit as $h$ approaches 0, not setting $h = 0$).
We have $\lim_{h
ightarrow0}\frac{1}{\sqrt{9 + h}+3}$.

Step4: Evaluate the limit

Substitute $h = 0$ into the expression $\frac{1}{\sqrt{9 + h}+3}$.
\[
\frac{1}{\sqrt{9+0}+3}=\frac{1}{3 + 3}=\frac{1}{6}
\]

Answer:

$\frac{1}{6}$