[Math] Taking limits with a square root in a quotient

Question

I'm trying to understand how to take the limit of

$$\lim_{x\to 0}\frac{\sqrt{36+x}-6}{x}$$

Wolfram alpha said I should use the l'hospital rule and take the derivative of the numerator and denominator but I don't understand why

Answer 1

$$\lim_{x\to 0}\frac{\sqrt{36+x}-6}{x}\frac{\sqrt{36+x}+6}{\sqrt{36+x}+6}=$$ $$=\lim_{x\to 0}\frac{36+x-36}{x(\sqrt{36+x}+6)}=\lim_{x\to 0}\frac{1}{\sqrt{36+x}+6}=1/12$$