Department of Energy Argonne National Laboratory Office of Science NEWTON's Homepage NEWTON's Homepage
NEWTON, Ask A Scientist!
NEWTON Home Page NEWTON Teachers Visit Our Archives Ask A Question How To Ask A Question Question of the Week Our Expert Scientists Volunteer at NEWTON! Frequently Asked Questions Referencing NEWTON About NEWTON About Ask A Scientist Education At Argonne L'Hopital's Rule Contradiction?
Name: Hassan S.
Status: student	
Age:  N/A
Location: N/A
Country: N/A
Date: N/A 

Hello, I approached our school's calculus teacher with the following question.
lim x -> infinity (x^2 + sin(x)) / (x^2).

It follows all the conditions for L'Hopital's rule, and when you work it out you get that the limit does not exist, but by graphing you can clearly tell that a limit does exist at one. Now if you split it up into lim x -> infinity (x^2/x^2) + lim x -> infinity (sin(x)), and simplify you can clearly see that it comes out to 1 + 0, which is one. Now the question is why does L'Hopital's rule give an incorrect answer. I do not have the derivation of L'Hopital's rule, so I cannot use that to see what is wrong, can you please tell me why L'Hopital's rule failed to produce a correct answer.

I think that you will find the following web site of particular interest!!!

For l'Hopital's rule to apply the limits of the numerator and denominator of the two functions:
f(x) and g(x) in the rational expression f(x)/g(x) must INDIVIDUALLY and
That is, BOTH lim(x--->a) [f(x)] = 0, or infinity and
SIMULTANEOUSLY lim(x--->a) [g(x)] = 0, or infinity. Otherwise you cannot
apply l'Hopital's rule.

If both conditions are met, it can be proven analytically that the
lim(x--->a) f(x)/g(x) =
lim(x--->a) f'(x)/g'(x) where f'(x) and g'(x) are the limits of the
functions f(x) and g(x) separately. If that ratio is still indeterminate,
then one applies the rule again and lim(x--->a) f'(x)/g'(x) =
lim(x--->a) f"(x)/g"(x) and one keeps applying the rule until the ratio is
no longer indeterminate, and that result.

Now what is happening in the case of x----> infinity for the rational
function: f(x)=(x^2 +sin(x))/x^2?
The limit of the numerator (x^2 +sin(x)) is indeterminate AND the limit
of the denominator g(x)=(x^2) also is. and f'(x)/g'(x) = (2*x +cos(x)) /
The limit of the numerator (2*x +cos(x)) and the denominator (2*x) as
x----> infinity are each still indeterminate, so apply l'Hopital's rule
The limit of the numerator is: f"'= (2 - sin(x)) IS indeterminate; but
g"'(x)=2 is NOT indeterminate.
So l'Hopital's rule does not apply as a method for obtaining the limit.

Closing point: One occasionally sees the rule spelled as above: (l'Hopital's rule) and as (l'Hospital's rule). I am not sure whether this is just an error or whether in Anglicizing the name there is an ambiguity about the spelling. In any case the rules are the same.

Vince Calder

The derivative of the numerator is infinite, as is the denominator. The first derivative yields
(2x+cos(x))/2x, still infinity over infinity. The next derivative yields
(2-sin(x))/2, now an
indeterminate function over 2. There is no way to identify the limit of
sin(x) as x-> infinity. As a result, L'Hopital's rule does not yield anything, neither a number nor infinity.
L'Hopital's rule requires that something result, a number or infinity. If
neither results, the
rule does not apply. In this case, L'Hopital's rule tells us nothing about
what to expect the
limit will be.

Dr. Ken Mellendorf
Physics Instructor
Illinois Central College

Click here to return to the Mathematics Archives

NEWTON is an electronic community for Science, Math, and Computer Science K-12 Educators, sponsored and operated by Argonne National Laboratory's Educational Programs, Andrew Skipor, Ph.D., Head of Educational Programs.

For assistance with NEWTON contact a System Operator (, or at Argonne's Educational Programs

Educational Programs
Building 360
9700 S. Cass Ave.
Argonne, Illinois
60439-4845, USA
Update: June 2012
Weclome To Newton

Argonne National Laboratory