Taking limits with a square root in a quotient 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
 A: $$\lim_{x\to 0}\frac{\sqrt{36+x}-6}{x}\frac{\sqrt{36+x}+6}{\sqrt{36+x}+6}=$$
$$=\lim_{x\to 0}\frac{36+x-36}{x(\sqrt{36+x}+6)}=\lim_{x\to 0}\frac{1}{\sqrt{36+x}+6}=1/12$$
A: $$\lim_{x\to 0}\frac{\sqrt{36+x}-6}{x}=\lim_{x\to 0}\frac{6}{x}(\sqrt{1+x/36}-1)=\lim_{x\to 0}\frac{6}{x}(1+x/72-1)=\frac{1}{12}$$
A: 
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?

Why? because it is of the form $0/0$ when you just naively try to put x as 0, and the derivative to both numerator and denominator exists:
$$\lim_{x\to 0}\frac{\sqrt{36+x}-6}{x}=\lim_{x\to0}\frac{d/dx(\sqrt{36+x}-6)}{d/dx(x)}=\lim_{x\to0}\frac{1/(2\sqrt{36+x})}{1}$$
Now you can just put the value of x as 0.
$$
\frac{1/(2\sqrt{36+x})}{1}|_{x=0}
=\frac{1}{2\times\sqrt{36+0}}
=\frac1{2\times6}=\frac1{12}$$

If you dont know about L'Hospital rule, it is used convert any indeterminate form usually $0/0,\infty/\infty$ into a determinate form when both numerator and denominator are diffrentiable as:
$$\lim_{x\to c}\frac{f(x)}{g(x)}=\lim_{x\to c}\frac{f'(x)}{g'(x)}$$
Sometimes but not always you may need to apply it successively like in:
$$\lim_{x\to0}\underbrace{\frac{\sin x-x}{x^3}}_{0/0}
=\lim_{x\to0}\underbrace{\frac{\cos x-1}{3x^2}}_{0/0}
=\lim_{x\to0}\underbrace{\frac{-\sin x}{6x}}_{0/0}
=\lim_{x\to0}\frac{-\cos x}{6}=-\frac16$$
