L'Hôpital's rule

L'Hôpital's rule in calculus uses derivatives in order to determine many otherwise hard to compute limits. The rule states: if you are trying to determine the limit of some quotient f(x)/g(x) and both the numerator and denominator approach 0 or infinity, then differentiate numerator and denominator and determine the limit of the quotient of the derivatives; if it exists, it will be the same as the original limit.

For example, a case of "0/0":

$\lim _{x\to 0}{\frac {\sin(x)}{x}}=\lim _{x\to 0}{\frac {\cos(x)}{1}}={\frac {1}{1}}=1$

and a case of "∞/∞":

$\lim _{x\to \infty }{\frac {\sqrt {x}}{\ln(x)}}=\lim _{x\to \infty }{\frac {\frac {1}{2{\sqrt {x}}}}{\frac {1}{x}}}=\lim _{x\to \infty }{\frac {\sqrt {x}}{2}}=\infty$

Sometimes, even limits which don't appear to be quotients can be handled with the same rule:

$\lim _{x\to \infty }x-{\sqrt {x^{2}-x}}=\lim _{x\to \infty }{\frac {\left(x+{\sqrt {x^{2}-x}}\right)\left(x-{\sqrt {x^{2}-x}}\right)}{x+{\sqrt {x^{2}+x}}}}=\lim _{x\to \infty }{\frac {x^{2}-(x^{2}-x)}{x+{\sqrt {x^{2}+x}}}}$

${\mbox{ }}=\lim _{x\to \infty }{\frac {x}{x+{\sqrt {x^{2}+x}}}}=\lim _{x\to \infty }{\frac {1}{1+{\frac {2x+1}{2{\sqrt {x^{2}+x}}}}}}={\frac {1}{1+1}}={\frac {1}{2}}$