L'Hopital's Rule Contradiction? ```Name: Hassan S. Status: student Age: N/A Location: N/A Country: N/A Date: N/A ``` 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. Replies: 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 Click here to return to the Mathematics Archives

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