question\nevaluate the limit: $lim_{x\rightarrow16}\frac{sqrt{x - 12}-2}{3x - 48}$\nanswer

Answer:

$\frac{1}{12}$

Explanation:

Step1: Rationalize the numerator

Multiply the fraction by $\frac{\sqrt{x - 12}+2}{\sqrt{x - 12}+2}$. [ \begin{align*} &\lim_{x\rightarrow16}\frac{\sqrt{x - 12}-2}{3x - 48}\times\frac{\sqrt{x - 12}+2}{\sqrt{x - 12}+2}\ =&\lim_{x\rightarrow16}\frac{(x - 12)-4}{(3x - 48)(\sqrt{x - 12}+2)}\ =&\lim_{x\rightarrow16}\frac{x - 16}{(3x - 48)(\sqrt{x - 12}+2)} \end{align*} ]

Step2: Factor the denominator

Factor out 3 from the denominator $3x - 48$ to get $3(x - 16)$. [ \begin{align*} &\lim_{x\rightarrow16}\frac{x - 16}{3(x - 16)(\sqrt{x - 12}+2)}\ =&\lim_{x\rightarrow16}\frac{1}{3(\sqrt{x - 12}+2)} \end{align*} ]

Step3: Substitute $x = 16$

[ \begin{align*} &\frac{1}{3(\sqrt{16 - 12}+2)}\ =&\frac{1}{3(2 + 2)}\ =&\frac{1}{12} \end{align*} ]