what is $lim_{x\rightarrow3}\frac{sqrt{x + 1}-2}{x - 3}$?\n0\n$\frac{1}{4}$\n4\ndne

Explanation:

Step1: Rationalize the numerator

Multiply the fraction by $\frac{\sqrt{x + 1}+2}{\sqrt{x + 1}+2}$. [ \begin{align*} &\lim_{x\rightarrow3}\frac{\sqrt{x + 1}-2}{x - 3}\times\frac{\sqrt{x + 1}+2}{\sqrt{x + 1}+2}\ =&\lim_{x\rightarrow3}\frac{(x + 1)-4}{(x - 3)(\sqrt{x + 1}+2)}\ =&\lim_{x\rightarrow3}\frac{x-3}{(x - 3)(\sqrt{x + 1}+2)} \end{align*} ]

Step2: Simplify the fraction

Cancel out the common factor $(x - 3)$ in the numerator and denominator. [ \begin{align*} &\lim_{x\rightarrow3}\frac{x-3}{(x - 3)(\sqrt{x + 1}+2)}\ =&\lim_{x\rightarrow3}\frac{1}{\sqrt{x + 1}+2} \end{align*} ]

Step3: Substitute $x = 3$

[ \begin{align*} &\frac{1}{\sqrt{3+1}+2}\ =&\frac{1}{\sqrt{4}+2}\ =&\frac{1}{2 + 2}\ =&\frac{1}{4} \end{align*} ]

Answer:

B. $\frac{1}{4}$