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