For what value $k$ is $f(x) = \begin{cases} \frac{\sqrt{2x+5}-\sqrt{x+7}}{x-2} & x \neq 2 \\ k & x = 2 \end{cases}$ continuous at $x=2$? For what value $k$ is the following function continuous at $x=2$?
$$f(x) = \begin{cases}
\frac{\sqrt{2x+5}-\sqrt{x+7}}{x-2} & x \neq 2 \\
k & x = 2
\end{cases}$$

All those square roots are weighing me down!  And $k$?  My mind's not where it's supposed to be today.  Thanks in advance for posting a solution!

We must have $f(2) = \lim_{x \to 2}(\frac{\sqrt{2x+5}-\sqrt{x+7}}{x-2})$ for $f(x)$ to be continuous at $x=2$. Since plugging in $2$ gives us $\frac{0}{0}$, we can use L'hopital's Rule and differentiate the top and bottom:
$$\lim_{x \to 2}(\frac{\sqrt{2x+5}-\sqrt{x+7}}{x-2})=\lim_{x \to 2}(\frac{2\frac{1}{2}(2x+5)^{-\frac{1}{2}}-\frac{1}{2}(x+7)^{-\frac{1}{2}}}{1})$$
$$\lim_{x \to 2}(\frac{2\frac{1}{2}(2x+5)^{-\frac{1}{2}}-\frac{1}{2}(x+7)^{-\frac{1}{2}}}{1})=\frac{1}{6}$$
So we see that we need $f(2)=\frac{1}{6}$, so $f(2) = k = \frac{1}{6}$.
