calculus - Two square roots in an indeterminate

Square roots are always causing trouble for me - especially finding the limit of a function when it's in the indeterminate form.

I got this far, but I don't know what to do next or even if it's right. Trying to go any further gives me my original problem.

This is the initial problem
$$\lim_{x\to5}\frac{\sqrt{ x^2+5}-\sqrt{30}}{(x-5)}$$

This is where I've gotten
$$\lim_{x\to5}\frac{\sqrt{x^2+5}-\sqrt{30}}{(x-5)} * \frac{\sqrt{x^2+5}+\sqrt{30}}{\sqrt{x^2+5}+\sqrt{30}} = \frac{x^2+5-30}{(x-5)\left(\sqrt{x^2+5}+\sqrt{30}\right)}$$

Answer

HINT $\ \ \ $ Factoring both of$\ \ \sqrt{x^2+5}^{\:2} - \sqrt{30}^{\:2}\ =\ x^2 - 5^2\ $ by difference of squares yields

$$\frac{\sqrt{x^2+5}-\sqrt{30}}{x-5}\ =\ \frac{x+5}{\sqrt{x^2+5}+\sqrt{30}}$$

Alternatively, if you're familiar with derivatives, note that the limit is $f\:'(5)$ for $f(x) = \sqrt{x^2+5}\:.$

Many limits can be simply calculated by recognizing them as instances of first derivatives and then calculating the derivatively rotely using known derivative rules. You can find a handful of examples in my prior posts starting here. Later in your course you will most likely encounter a generalization of this observation known as l'Hôpital's Rule.