Question

What is L'Hôpital's rule used for?

Answer 1

L'Hôpital's rule is a really useful tool for evaluating limits of an indeterminate form `0/0`, or, `oo/oo`.

The theorem states that if we have a limit of an indeterminate form `0/0`, or, `oo/oo`, then:

`lim_(x rarr a) f(x)/g(x) = lim_(x rarr a) (f'(x))/(g'(x))`

Providing the limit does actually exist.

Proof

A specific proof for the case `0/0` is fairly easy, if `f` and `g` are differentible at `a` and the derivatives are continuous at `a`.

In this case, `f(a)=g(a)=0`:

`lim_(x rarr a) f(x)/g(x) = lim_(x rarr a) (f(x)-0)/(g(x)-0)`

`" " = lim_(x rarr a) (f(x)-f(a))/(g(x)-g(a))`

`" " = lim_(x rarr a) ((f(x)-f(a))/(x-a)) / ((g(x)-g(a))/(x-a)`

`" " = (lim_(x rarr a) (f(x)-f(a))/(x-a)) / (lim_(x rarr a)(g(x)-g(a))/(x-a)`

`" " = (f'(a)) / (g'(a))`

`" " = lim_(x rarr a) (f'(x)) / (g'(x))` QED

Example

`L = lim_(x rarr 0) (e^x-1)/x`

If we put `x=0` we get an indeterminate form `0/0`, so we can apply L'Hôpital's rule to get:

`L = lim_(x rarr 0) (d/dx (e^x-1))/(d/dx x`
`\ \ = lim_(x rarr 0) (e^x)/1`
`\ \ = lim_(x rarr 0) (e^x)/1`
`\ \ = 1/1`
`\ \ = 1`