evaluate the limit below in two steps by using algebra to simplify the difference quotient and then evaluating the limit. $lim_{h
ightarrow0^{+}}left(\frac{sqrt{h^{2}+9h + 2}-sqrt{2}}{h}
ight)=lim_{h
ightarrow0^{+}}left(square
ight)=square$

Question

Accepted Answer

# Explanation:
## Step1: Rationalize the numerator
Multiply the fraction by $\frac{\sqrt{h^{2}+9h + 2}+\sqrt{2}}{\sqrt{h^{2}+9h + 2}+\sqrt{2}}$.
$$
\begin{align*}
&\lim_{h
ightarrow0^{+}}\frac{\sqrt{h^{2}+9h + 2}-\sqrt{2}}{h}\times\frac{\sqrt{h^{2}+9h + 2}+\sqrt{2}}{\sqrt{h^{2}+9h + 2}+\sqrt{2}}\
=&\lim_{h
ightarrow0^{+}}\frac{(h^{2}+9h + 2)-2}{h(\sqrt{h^{2}+9h + 2}+\sqrt{2})}\
=&\lim_{h
ightarrow0^{+}}\frac{h^{2}+9h}{h(\sqrt{h^{2}+9h + 2}+\sqrt{2})}
\end{align*}
$$
## Step2: Simplify the fraction and evaluate the limit
Cancel out the common - factor $h$ in the numerator and denominator.
$$
\begin{align*}
&\lim_{h
ightarrow0^{+}}\frac{h^{2}+9h}{h(\sqrt{h^{2}+9h + 2}+\sqrt{2})}\
=&\lim_{h
ightarrow0^{+}}\frac{h(h + 9)}{h(\sqrt{h^{2}+9h + 2}+\sqrt{2})}\
=&\lim_{h
ightarrow0^{+}}\frac{h + 9}{\sqrt{h^{2}+9h + 2}+\sqrt{2}}
\end{align*}
$$
Substitute $h = 0$ into the expression: $\frac{0 + 9}{\sqrt{0^{2}+9\times0+2}+\sqrt{2}}=\frac{9}{2\sqrt{2}}=\frac{9\sqrt{2}}{4}$.

# Answer:
$\frac{9\sqrt{2}}{4}$

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QUESTION IMAGE

Question

Explanation:

Step1: Rationalize the numerator

Step2: Simplify the fraction and evaluate the limit

Answer: