Answer
$\frac{1}{4}$
Work Step by Step
\begin{align*}
\lim _{x \rightarrow 3}\frac{\sqrt{x+1}-2}{x-3}&=\lim _{x \rightarrow 3}\frac{\sqrt{x+1}-2}{x+1-4}\\
&=\lim _{x \rightarrow 3}\frac{\sqrt{x+1}-2}{(\sqrt{x+1}-2)(\sqrt{x+1}+2)}\\
&=\lim _{x \rightarrow 3}\frac{1}{\sqrt{x+1}+2}\\
&=\frac{1}{\sqrt{3+1}+2}\\
&=\frac{1}{4}.
\end{align*}
