[h2]Quotient Rule[/h2] $$\lim_{x\to c} \frac{f(x)}{g(x)} = \frac{\lim_{x\to c} f(x)}{\lim_{x\to c} g(x)}$$ [h2]L'Hopital's rule[/h2] $$\lim_{x\to c} \frac{f(x))}{g(x))} = \lim_{x\to c} \frac{f'(x))}{g'(x))}$$ [h3]Proof[/h3] Let f(c) = g(c) = 0 $$\lim_{x\to c} \frac{f(x))}{g(x))} = \lim_{x\to c} \frac{f(x) - 0}{g(x) - 0}$$ Then, due to the above assumption of the value c: $$=\lim_{x\to c} \frac{f(x)-f(c)}{g(x)-g(c)}$$ And then if we multiply by 1/(x-c): $$=\lim_{x\to c} \frac{(\frac{f(x)-f(c)}{x-c})}{(\frac{g(x)-g(c)}{x-c})}$$ Then by the Quotient Rule from above: $$=\frac{\lim_{x\to c} (\frac{f(x)-f(c)}{x-c})}{\lim_{x\to c} (\frac{g(x)-g(c)}{x-c})}$$ Then this appears like [delta]y/[delta]x at the top and bottom of the fraction, which is exactly the form of the derivatives of both functions (for the variable c), and so we get: $$=\frac{f'(c)}{g'(c)}$$ And hence: $$\lim_{x\to c} \frac{f(x)}{g(x)}=\lim_{x\to c} \frac{f'(x)}{g'(x)}$$