## Calculus (3rd Edition)

$$\frac{1}{80}.$$
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 $$\lim _{x \rightarrow 3} \frac{1/(2\sqrt{x+1})}{3x^{2}-7}=\frac{1/4}{27-7}=\frac{1}{80}.$$