Finding Limits Using Natural Logs and L'Hopital's RuleDate: 05/03/98 at 20:12:18 From: Brook Bailey Subject: L'Hospital's rule Use L'Hopital's Rule to find the following limits: lim (x goes to 0) [(cos(2x))^(3/(x^2))] I know that if I use the chain rule, the equation gets very messy. I think that I need to use the e function or bring a ln (the natural logarithm function) into the picture, but I am not quite sure how. Please help me. Date: 05/04/98 at 16:34:13 From: Doctor Sam Subject: Re: L'Hospital's rule Brook, You are quite correct. L'Hopital's Rule only applies to functions that are quotients that approach the indeterminate form 0/0 or infinity/infinity. This function approaches the indeterminate form 1^infinity. Here is the idea. Either lim f(x)^g(x) approaches a limit or it does not. If it does, then lim f(x)^g(x) = L. In this case, if we take natural logarithms of both sides of the equation, we get: ln [lim f(x)^g(x)] = ln (L) Now, the natural logarithm function is continuous. One of the properties of continuous functions is that you can "bring them inside a limit." For example: [lim sqrt(x)]^2 as x -> 3 = lim [sqrt(x)]^2 = lim x = 3. Since "squaring" is continuous, we can bring it inside the limit sign. So in general, when you have a limit, like lim f(x)^g(x) = L, we can "take logs" to get: ln [lim f(x)^g(x)] = ln (L) lim ln [f(x)^g(x)] = ln (L) lim g(x) ln [f(x)] = ln (L) using a property of logarithms to simplify the expression, namely that ln(a^b) = b ln(a). Notice that the result of this gives ln(L) -- the logarithm of the original limit. In your problem, if we take logarithms, we get: lim [(cos (2x))^(3/(x^2))] as x -> 0 ln [lim [(cos (2x))^(3/(x^2))]] as x -> 0 lim [ln [(cos (2x))^(3/(x^2))]] as x -> 0 lim (3/x^2) ln (cos (2x)) as x -> 0 Finally, we have a quotient: 3 ln (cos (2x)) lim ---------------- as x -> 0 x^2 Note that cos(2x) -> 1 and ln(1) = 0, so the numerator approaches zero. The denominator also approaches zero. These two conditions mean that L'Hopital's Rule applies. L'Hopital's Rule states that this limit, if it exists, is the same as the limit of the ratio of the derivatives of the numerator and denominator. So: 3*ln (cos (2x)) -6sin(2x)/cos(2x) lim ---------------- = lim ----------------- x^2 2x I used the Chain Rule to find the derivative of 3*ln(cos(2x)). Since it is awkward to keep a fraction in the numerator of another fraction, I am going to simplify this to: -6sin(2x) lim ----------- 2xcos(2x) As x ->0, the numerator approaches 0 and the denominator still approaches 0. That means that we can apply L'Hopital's Rule a second time: -6sin(2x) -12cos(2x) lim ----------- = lim --------------------- 2xcos(2x) -4xsin(2x) + 2cos(2x) I needed the product rule to find the derivative of 2xcos(2x). Now what happens as x ->0? The numerator approaches -12 and the denominator approaches 2. The limit is, therefore, -12/2 = -6. But we are not quite done. Remember that we took the logarithm of your original function. We said that if we could find its limit that this would be ln (L) -- the logarithm of the original limit. So now we have to solve the equation: ln (L) = -6 Use both sides of the equation as exponents of the exponential function e: e^[ln L] = e^-6 One of the basic properties of exponential functions and logarithm functions is that they undo each other, that is, they are inverses. So e^ln(L) = L. Therfore, L = e^-6 or L = 1/e^6. I hope that helps. -Doctor Sam, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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