Answer
$$\frac{1}{80}.$$
Work Step by Step
Since we have
$$
\lim _{x \rightarrow 3} \frac{\sqrt{x+1}-2}{x^{3}-7x-6}=\frac{2-2}{27-21-6}=\frac{0}{0}
$$
then we can apply L’Hôpital’s Rule as follows
$$
\lim _{x \rightarrow 3} \frac{1/(2\sqrt{x+1})}{3x^{2}-7}=\frac{1/4}{27-7}=\frac{1}{80}.
$$
