28. evaluate $lim_{x \to 2} \frac{sqrt{x + 7}-3}{x - 2}$

Explanation:

Step1: Rationalize the numerator

Multiply the fraction by $\frac{\sqrt{x + 7}+3}{\sqrt{x + 7}+3}$. [ \begin{align*} &\lim_{x\rightarrow2}\frac{\sqrt{x + 7}-3}{x - 2}\times\frac{\sqrt{x + 7}+3}{\sqrt{x + 7}+3}\ =&\lim_{x\rightarrow2}\frac{(\sqrt{x + 7})^2-3^2}{(x - 2)(\sqrt{x + 7}+3)}\ =&\lim_{x\rightarrow2}\frac{x+7 - 9}{(x - 2)(\sqrt{x + 7}+3)}\ =&\lim_{x\rightarrow2}\frac{x - 2}{(x - 2)(\sqrt{x + 7}+3)} \end{align*} ]

Step2: Simplify the fraction

Cancel out the common factor $(x - 2)$ in the numerator and denominator. [ \begin{align*} &\lim_{x\rightarrow2}\frac{x - 2}{(x - 2)(\sqrt{x + 7}+3)}\ =&\lim_{x\rightarrow2}\frac{1}{\sqrt{x + 7}+3} \end{align*} ]

Step3: Substitute $x = 2$

[ \begin{align*} &\frac{1}{\sqrt{2+7}+3}\ =&\frac{1}{\sqrt{9}+3}\ =&\frac{1}{3 + 3}\ =&\frac{1}{6} \end{align*} ]

Answer:

$\frac{1}{6}$