# High School Calculus/L'Hôpital's Rule

### L'Hôpital's Rule

This is a really fun rule to deal with. As explained in a different section the way you have learned to deal with fractional limits is by factoring. L'Hôpital's Rule is a much easier way to deal with these fractional limit functions.

When you use this rule you must first prove that you get an indeterminate form when evaluating the limit.

Indeterminate form can come in a variety of different ways.

Indeterminate forms: ${\displaystyle {\frac {0}{0}}}$, ${\displaystyle {\frac {\infty }{\infty }}}$, ${\displaystyle 0*\infty }$, ${\displaystyle {\frac {-\infty }{-\infty }}}$, ${\displaystyle \infty -\infty }$, ${\displaystyle {\frac {-\infty }{\infty }}}$, ${\displaystyle {\frac {\infty }{-\infty }}}$, ${\displaystyle 0^{0}}$, ${\displaystyle 1^{\infty }}$, ${\displaystyle \infty ^{0}}$

There are a lot of them, but the more you work with them and become familiar with these indeterminate forms they will be committed to memory.

Ex: 1

Show that the limit is in an indeterminate form

${\displaystyle \lim _{x\to 0}{\frac {x^{2}}{x}}}$

The next step to L'Hôpital's Rule is to look at the fraction as ${\displaystyle {\frac {f(x)}{g(x)}}}$

As long as ${\displaystyle g(x)\neq 0}$ you can use L'Hôpital's Rule.

The formula for L'Hôpital's Rule is ${\displaystyle \lim _{x\to c}{\frac {f(x)}{g(x)}}}$ as long as f and g are differentiable on the open interval (a,b)

Ex: 2

Determine if you can use L'Hôpital's Rule. If you can then evaluate the limit using L'Hôpital's Rule.

${\displaystyle \lim _{x\to 0}{\frac {3x^{3}-x}{x-5}}}$

You will notice when trying to evaluate some limits using L'Hôpital's Rule that the function just switches back and forth. When this happens you will probably have multiply a variable into the function, as long as it equals one.

Ex: 3

Determine if you can use L'Hôpital's Rule. If you can then evaluate the limit using L'Hôpital's Rule

${\displaystyle \lim _{x\to \infty }{\frac {\sqrt {x}}{\sqrt {x+1}}}}$

Solution to Ex: 3

${\displaystyle \lim _{x\to \infty }{\frac {\sqrt {x}}{\sqrt {x+1}}}={\frac {\infty }{\infty }}}$

${\displaystyle \lim _{x\to \infty }{\frac {\frac {1}{2*{\sqrt {x}}}}{\frac {1}{2*{\sqrt {x+1}}}}}=\lim _{x\to \infty }{\frac {\sqrt {x+1}}{\sqrt {x}}}}$

${\displaystyle \lim _{x\to \infty }{\frac {\frac {1}{2*{\sqrt {x+1}}}}{\frac {1}{2*{\sqrt {x}}}}}=\lim _{x\to \infty }{\frac {\sqrt {x}}{\sqrt {x+1}}}}$

As you can see we are getting no where fast. Something else needs to be done

${\displaystyle \lim _{x\to \infty }{\frac {\sqrt {x}}{\sqrt {x+1}}}*{\frac {\frac {1}{x^{2}}}{\frac {1}{x^{2}}}}}$

${\displaystyle \lim _{x\to \infty }{\frac {\sqrt {\frac {x}{x}}}{\sqrt {{\frac {x}{x}}+{\frac {1}{x}}}}}}$

${\displaystyle \lim _{x\to \infty }{\frac {1}{\sqrt {1+{\frac {1}{x}}}}}}$

${\displaystyle {\frac {1}{\sqrt {1+{\frac {1}{\infty }}}}}}$

${\displaystyle {\frac {1}{\sqrt {1+0}}}=1}$

${\displaystyle L=1}$