L'Hopital's Rule ```Name: Brian Status: student Age: 18 Location: N/A Country: N/A Date: N/A ``` Question: My Calculus teacher made up quiz off the top of his head for a section on L'Hopital's Rule. The last question was to find the limit as x approaches infinity in the equation cos(x)/ln(x). When you try to use L'Hopital's you don't get an answer, yet when you graph the equation, you see that it approaches 0. My teacher was not able to explain why L'Hopital's rule falls flat on its face here. Can you tell us how to solve the problem and why L'Hopital's rule does not work? Replies: L'Hospital's Rule does work in this case. All the rule claims is that lim f(x)/g(x) = lim f'(x)/g'(x) which is true. Unfortunately, this information doesn't help you find the limit. To do that in this case, the best tool is common sense: cos(x) is bounded; ln(x) is not. This is a great lesson whose relevance goes way beyond math. L'Hospital's rule is not for finding limits. It's just a statement about functions and their derivatives, and it doesn't care what you want to prove. Tim Mooney I believe that the rule is applicable when the denominator approaches zero. In this case ln(x) goes to -infinity, as x approaches zero. Dr. Myron Hello, L'Hopital's rule is still applies as follows: Lim(x -->0) of [cosx/lnx] = -lim(x -->0) of [(sinx)/(1/x)] = -lim (x -->0) [x sinx] =0. L'Hopital's rule is used in cases where both the numerator and the denominator go to zero at some limits. In the present case, for x -> infinity, L'Hopital's rule does not apply because: Lim(x -->infinity) of [cosx/lnx] = ~ cosx/infinity =0. Cosine of x has values between -1 to +1 and when divided by a large number approaches zero. AK Dr. Ali Khounsary Advanced Photon Source Argonne National Laboratory Click here to return to the Mathematics Archives

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