Ask A Scientist Mathematics Archive

name         Hassan S.
status       student
age          16

Question -   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!!!
http://www.dougshaw.com/findtheerror/FTELHopital.html

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 SIMULTANEOUSLY be 
indeterminate.
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)) /
(2*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
again:
    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
=====================================================