what is $lim_{x\rightarrow5}\frac{sqrt{x + 4}-3}{x - 5}$?\n- $\frac{1}{6}$\n0\n$\frac{1}{6}$\ndne

Explanation:

Step1: Rationalize the numerator

Multiply the fraction by $\frac{\sqrt{x + 4}+3}{\sqrt{x + 4}+3}$. [ \begin{align*} &\lim_{x\rightarrow5}\frac{\sqrt{x + 4}-3}{x - 5}\times\frac{\sqrt{x + 4}+3}{\sqrt{x + 4}+3}\ =&\lim_{x\rightarrow5}\frac{(\sqrt{x + 4})^2-3^2}{(x - 5)(\sqrt{x + 4}+3)}\ =&\lim_{x\rightarrow5}\frac{x + 4-9}{(x - 5)(\sqrt{x + 4}+3)}\ =&\lim_{x\rightarrow5}\frac{x - 5}{(x - 5)(\sqrt{x + 4}+3)} \end{align*} ]

Step2: Simplify the fraction

Cancel out the common factor $(x - 5)$ in the numerator and denominator. [ \begin{align*} &\lim_{x\rightarrow5}\frac{x - 5}{(x - 5)(\sqrt{x + 4}+3)}\ =&\lim_{x\rightarrow5}\frac{1}{\sqrt{x + 4}+3} \end{align*} ]

Step3: Substitute $x = 5$

[ \begin{align*} &\frac{1}{\sqrt{5+4}+3}\ =&\frac{1}{\sqrt{9}+3}\ =&\frac{1}{3 + 3}\ =&\frac{1}{6} \end{align*} ]

Answer:

C. $\frac{1}{6}$