L'Hopital's Rule- Sean Dandeneau

L’Hopital’s Rule

We recently learned how to use L’Hopital’s rule when we must do so. This situation appears whenever we try to evaluate limits that produce indeterminate forms when we try direct substitution.

To use L’Hopital’s rule, you must find the derivatives of the numerator and the denominator. Once this is found, you plug in c for  x and solve for the limit as x->c.

Lim of f(x)/g(x) as x->c = lim of f’(x)/g’(x) as x->c = lim of f’’(x)/g’’(x) as x->c

Lim 1-cos(x)/x as x->0

= 0/0

    So we must use L’Hopital’s Rule…

Lim 0-(-sin(x))/1 as x->0

= lim sin(x) as x->0

= sin(0)

= 0

When doing L’Hopital’s Rule, we must keep in mind that all derivative rules apply. We can find as many derivatives as we need to in order to solve for the limit at c.

Another example:

f(x)=lnx^5/x

Lim x->infinity lnx^5/x

    = lim x->infinity 5lnx/x

    = lim x->infinity (5/x)/1

    = lim x-> infinity 5/x

    = lim x-> infinity 0/1

    = 0