Question

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

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:

$\frac{9\sqrt{2}}{4}$