Calculus - Solving limits with square roots I am having trouble understanding how to solve this limit by rationalizing. I have the problem correct (I used Wolfram Alpha of course), but I still don't understand how it is completed. I was trying to solve this by multiplying both the numerator and the denominator by $\sqrt{x^2+11}+6$, but I am stuck. Is that the way to solve this? If so, could you walk me through it so I can solve others?
Problem:


$$\begin{align}
\lim_{x\to 5}\frac{\sqrt{x^2+11}-6}{x-5} 
  &= \lim_{x\to 5}\frac{\left(\sqrt{x^2+11}-6\right)
    \left(\sqrt{x^2+11}+6\right)}{(x-5)\left(\sqrt{x^2+11}+6\right)} \\[2ex]
  &= \lim_{x\to 5}\frac{\left(\sqrt{x^2+11}\right)^2-6^2}
    {(x-5)\left(\sqrt{x^2+11}+6\right)} \\[2ex]
  &= \lim_{x\to 5}\frac{x^2+11-36}{(x-5)\left(\sqrt{x^2+11}+6\right)} \\[2ex]
  &= \lim_{x\to 5}\frac{x^2-25}{(x-5)\left(\sqrt{x^2+11}+6\right)} \\[2ex]
  &= \lim_{x\to 5}\frac{(x-5)(x+5)}{(x-5)\left(\sqrt{x^2+11}+6\right)} \\[2ex]
  &= \lim_{x\to 5}\frac{x+5}{\sqrt{x^2+11}+6} \\[2ex]
  &= \frac{5+5}{\sqrt{5^2+11}+6} \\[2ex]
  &= \frac{10}{12} \\[2ex]
  &= \frac{5}{6}
\end{align}$$
