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# 3 8x 2 x4 2x 1 x we shall now discuss the limit x 100

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Unformatted text preview: q log x  x log 1  q log x  x log x x 1  q log x  x log x q log x x log x Now given any sufficiently large number x, we can apply the mean value theorem to the function log on the interval x q log x, log x to obtain a number f x between x q log x and log x at which the derivative of the function log is equal to log x log x q log x x x q log x and we can express f x in the form x g x where 0 g x q log x. Thus log x log x q log x 1  q log x x gx and we conclude that q log x q log x  x log x q log x x log x  q log x x x gx  q log x 1 x gx x Therefore q log x x q log x  x log 1  q log x q log x x x gx x x q log x 1 1 log x 2 q2 x q 2 log x 2  log x x q log x 1 qx Ý. This completes our consideration of the case q  0.  and the latter expression approaches 0 as x Finally we consider the case q  0. We write p  q. Then p  0 and we want to show that p log x x lim 1 1   1. p x xÝx We need to show that p log x x log 1  p log x 0 x as x Ý. Now whenever x is sufficiently large we have p log x x  p log x x log 1  p log x  x log x x and for some number g x satisfying 0 gx  x log x  p log x p log x we have 254 p log x x log x p log x x log 1  p log x x px log x xg x p log x  p log x and so p log x x p log x  p log x 1 x xg x p log x x log 1  1 x x  p log x  log x 2 x log x p x p 1 Ý. and the latter expression approaches 0 as x 6. Suppose that we want to evaluate the limit 3 . 1  x  x2 Would it be correct to use L’Hôpital’s rule by taking f x  3 and g x  1  x  x 2 for every x and then arguing that fx fx 3 lim  lim  lim  lim 0  0? x Ý 1  x  x2 xÝgx xÝg x x Ý 1  2x The answer is certainly correct, but is the reasoning correct? Have we made a valid use of L’Hôpital’s rule? lim xÝ Solution: Yes, we have made a valid application of L’Hôpital’s rule. The fact that the denominator of the fraction 3 1  x  x2 approaches Ý as x Ý is all we need; even though most elementary calculus books give the impression that L’Hôpital’s rule requires both the numerator and the denominator to approach Ý. 7. Evaluate the limits sin x lim e x xÝ and sin x lim xe . x Ý log x Should L’Hôpital’s rule be used to evaluate these limits? L’Hôpital’s rule would say that sin x e sin x cos x lim e x  lim xÝ xÝ 1 provided that the latter limit exists. But it doesn’t. So L’Hôpital’s rule can’t be used here. Note, however that sin x lim e x  0. xÝ L’Hôpital’s rule would say that sin x e sin x  x e sin x cos x lim xe  lim 1 x Ý log x xÝ x and the latter limit does not exist. Nor does the original limit. In fact, the graph sin x y  xe log x is 255 30 25 20 15 10 5 10 20 x 30 40 50 10 The Exponential and Logarithmic Functions Some Exercises on the Exponential and Logarithmic Functions 1. Given that a is a positive number and that f x  a x for every real number x, prove that f x  a x log a for every number x. In other words, prove that the number we called  a in the preceding sections is ju...
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## This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

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