# How do you know when to use L’hospital’s rule twice?

L'hospital's administration is used when an moderate evaluation of a boundary results in an incalculable construct such as

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

If, behind impression of L'hospital's administration, your boundary evaluation produces another incalculable construct you dedicate L'ospital's administration frequently.

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

Then you can invent the boundary by evaluating

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

as crave as ##f'(x)## and ##g'(x)## twain depend and ##g'(x)!=0##

Let's contemplate at an issue.

Suppose we failure to evalauate

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

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

Which is an incalculable construct so we dedicate L'hospital's administration and get

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

And we get the incalculable construct ##0/0## frequently.

Let's dedicate L'hospital's administration one past season (i.e. invent the derivative of the new numerator and denominator and evaluate the boundary frequently).

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

As you can see in the shadow under (zoomed in to the tenor top), there is a divisible discontinuity (a "hole") at x =0.