## Calculus (3rd Edition)

$$\frac{1}{4}$$
From the figure, we see that $$\lim _{x \rightarrow 4}\left(\frac{1}{\sqrt{x}-2}-\frac{4}{x-4}\right) =\frac{1}{4}$$ and algebraically, we have \begin{aligned} \lim _{x \rightarrow 4}\left(\frac{1}{\sqrt{x}-2}-\frac{4}{x-4}\right) &=\lim _{x \rightarrow 4} \frac{1}{\sqrt{x}+2} \\ &=\frac{1}{\sqrt{4}+2}\\ &=\frac{1}{4} \end{aligned}