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:
2 Answers
$\begingroup$$$\begin{align} \lim_{x\to 5}\frac{\sqrt{x^2+11}-6}{x-5} &= \lim_{x\to 5}\frac{\left(\sqrt{x^2+11}-6\right) \left(\sqrt{x^2+11}+6\right)}{(x-5)\left(\sqrt{x^2+11}+6\right)} \\[2ex] &= \lim_{x\to 5}\frac{\left(\sqrt{x^2+11}\right)^2-6^2} {(x-5)\left(\sqrt{x^2+11}+6\right)} \\[2ex] &= \lim_{x\to 5}\frac{x^2+11-36}{(x-5)\left(\sqrt{x^2+11}+6\right)} \\[2ex] &= \lim_{x\to 5}\frac{x^2-25}{(x-5)\left(\sqrt{x^2+11}+6\right)} \\[2ex] &= \lim_{x\to 5}\frac{(x-5)(x+5)}{(x-5)\left(\sqrt{x^2+11}+6\right)} \\[2ex] &= \lim_{x\to 5}\frac{x+5}{\sqrt{x^2+11}+6} \\[2ex] &= \frac{5+5}{\sqrt{5^2+11}+6} \\[2ex] &= \frac{10}{12} \\[2ex] &= \frac{5}{6} \end{align}$$
$\endgroup$ 1 $\begingroup$A possible step-by-step solution: write $x=y+5$ (so that you are looking for a limit as $y\to 0$), and the denominator is $x-5=y$
$$\begin{align} \sqrt{x^2+11} &= \sqrt{(y+5)^2+11} = \sqrt{y^2+10y + 36} = \sqrt{36}\sqrt{1+\frac{10}{36}y+\frac{y^2}{36}} \\ &= 6 \sqrt{1+\frac{5}{18}y+\frac{y^2}{36}} \end{align} $$
From there,
$$\begin{align} \sqrt{x^2+11} - 6 &= 6 \left( \sqrt{1+\frac{5}{18}y+\frac{y^2}{36}} - 1\right) \end{align} $$ Now, if you know Taylor series, you can immediately apply the fact that $\sqrt{1+u} = 1+\frac{u}{2} + o(u)$ when $u\to 0$ to get that $$\begin{align} \sqrt{x^2+11} - 6 &= 6 \left( 1+\frac{5}{36}y+ o(y) - 1\right) = 6 \left( \frac{5}{36}y+ o(y)\right) = \frac{5}{6}y+ o(y) \end{align} $$ and now $$\frac{\sqrt{x^2+11} - 6}{y} = \frac{\frac{5}{6}y+ o(y)}{y} = \frac{5}{6} + o(1) \xrightarrow[y\to 0]{} \frac{5}{6}.$$
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