evaluate the limit below in two steps by using algebra to simplify the difference quotient and then…

evaluate the limit below in two steps by using algebra to simplify the difference quotient and then evaluating the limit. $lim_{h\rightarrow0^{+}}left(\frac{sqrt{h^{2}+9h + 2}-sqrt{2}}{h}\right)=lim_{h\rightarrow0^{+}}left(square\right)=square$
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\rightarrow0^{+}}\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\rightarrow0^{+}}\frac{(h^{2}+9h + 2)-2}{h(\sqrt{h^{2}+9h + 2}+\sqrt{2})}\ =&\lim_{h\rightarrow0^{+}}\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\rightarrow0^{+}}\frac{h^{2}+9h}{h(\sqrt{h^{2}+9h + 2}+\sqrt{2})}\ =&\lim_{h\rightarrow0^{+}}\frac{h(h + 9)}{h(\sqrt{h^{2}+9h + 2}+\sqrt{2})}\ =&\lim_{h\rightarrow0^{+}}\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:
$\frac{9\sqrt{2}}{4}$