L’hospital’s rule is used when an initial evaluation of a limit results in an indeterminate form such as

##0/0## or ##(+-oo)/(+-oo)##

If, after application of L’hospital’s rule, your limit evaluation produces another indeterminate form you apply L’ospital’s rule again.

L’hospital’s rule states the if

##lim_(x->c)(f(x))/(g(x))=0/0 or (+-oo)/(+-oo)##

Then you can find the limit by evaluating

##lim_(x->c)(f'(x))/(g'(x))##

as long as ##f'(x)## and ##g'(x)## both exist and ##g'(x)!=0##

Let’s look at an example.

Suppose we want to evalauate

##lim_(x->0)(cos x – 1)/(x^2)##

By direct substitution we get

##lim_(x->0)(cos x – 1)/(x^2)=(cos 0 – 1)/(0^2)=(1-1)/0=0/0##

Which is an indeterminate form so we apply L’hospital’s rule and get

##lim_(x->0)(-sin x)/(2x)=(-sin 0)/(2*0)=0/0##

And we get the indeterminate form ##0/0## again.

Let’s apply L’hospital’s rule one more time (i.e. find the derivative of the new numerator and denominator and evaluate the limit again).

##lim_(x->0)(-sin x)/(2x)=lim_(x->0)(-cos x)/2=(-1/2)##

As you can see in the image below (zoomed in to the problem point), there is a removable discontinuity (a “hole”) at x =0.