By substituting into these continuous functions we once again see that this is a limit of the form \(\frac{0}{0}\). So we take the usual approach of expanding the expression using the identity \(a^2-b^2=(a-b)(a+b)\) and we get
\[\begin{align*} \frac{\sqrt{x+13}-2\sqrt{x+1}}{x^2-9}&=\frac{\sqrt{x+13}-2\sqrt{x+1}}{x^2-9} \frac{\sqrt{x+13}+2\sqrt{x+1}}{\sqrt{x+13}+2\sqrt{x+1}}\\ &=\frac{(x+13)-4(x+1)}{x^2-9}\frac{1}{\sqrt{x+13}+2\sqrt{x+1}}\\ &=\frac{-3(x-3)}{(x-3)(x+3)}\frac{1}{\sqrt{x+13}+2\sqrt{x+1}} .\\ \end{align*}\]
After simplifying we have continuous functions which we can substitute into, and the limit turns out to be
\(\displaystyle \lim_{x\to 3}\frac{\sqrt{x+13}-2\sqrt{x+1}}{x^2-9} =\lim_{x\to 3}\frac{-3}{x+3}\frac{1}{\sqrt{x+13}+2\sqrt{x+1}} =\frac{-3}{6}\frac{1}{\sqrt{16}+2\sqrt{4}}=\frac{-1}{16}\ . \)
