1873_solutions

# And 0 j n we shall define binomial coefficients more

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Unformatted text preview: case   2 we approached the problem using L’Hôpital’s rule and again using the mean value theorem. This 251 time we shall use only the mean value theorem. For each x  0 we choose a number f x between x and x  1 such that  log f x  1 log x  1  log x   . fx If we define log u  1 gu  u for u  0 then since  1 log u  2 log u  1  1 log u  log u  2 0 gu u2 u2 whenever u  exp  1 we have  log x  1  log x  1  1 log x  1  log x  x x1 whenever x  exp  1 and so  lim log x  1 xÝ  log x  0. Now we discuss the limit lim xÝ log x  1 log x x  lim exp x log xÝ log x  1 log x  lim exp x log log x  1 xÝ log log x We look first at the limit lim x log log x  1 xÝ  lim xÝ log log x log log x  1 log log x 1 x Once again, L’Hôpital’s rule can be used but the mean value theorem presents an attractive alternative. For each x  0 we choose a number f x between x and x  1 such that 1 log log x  1 log log x  f x log f x from which we see (in the same way we argued in the preceding exercises) that log log x  1 log log x x x 1 x log x x  1 log x  1 x and so log log x  1 lim xÝ log log x 1 x 0 and we conclude that lim xÝ Now we consider the limit log x  1 lim xÝ log x x log x log x  1 log x x  exp 0  1.  lim exp xÝ x log x log  lim exp xÝ x log x log x  1 log x log log x  1 log log x We look first at the limit lim xÝ x log x log log x  1 log log x 252  lim xÝ log log x  1 1 x log x log log x Arguing as above we see that x log x x log x log log x  1 log log x x log x x  1 log x  1 1 x log x and so log log x  1 lim xÝ log log x 1 1 x log x and we conclude that log x  1 log x lim xÝ 4. a. Prove that if 0  a  b and x x log x  exp 1  e. 1 then bx ax xb x 1 . ba Suppose that x 1. For every positive number t we define f t  tx and we apply the mean value theorem to f on the interval a, b . xa x 1 b. Prove that lim log x  1 xÝ log x log x  Ý. log x Solution: From part a we know that, whenever x is sufficiently large, log x log x log x  1 log x log x  1 log x 1 log x log x log x log x log x  1 log x 1 Thus log x  1 log x  1 log x log x log x log x log x  1 x log x log x  1 log x log x log x log x log x log x . To see that the latter expression clearly approaches Ý as x Ý, we write it as log x  1 log x exp log x log log x . x 1 x It is easy to see that lim xÝ log x  1 log x 1 1 x and since log log x  2 for all sufficiently large x we have exp log x log log x exp 2 log x x x x for all x sufficiently large and so lim xÝ exp log x log log x x  Ý. c. Prove that lim log x  1 xÝ log log x log x log log x  0. 5. Prove that if q is any given number then lim x q 1 xÝ q log x x x 1 Solution: The assertion we want to prove is obviously true when q  0. We now consider the case 253 q  0. We need to show that q log x x q log x  x log 1 as x 0 Ý. Since lim xÝ log x x 0 we know that q log x 0 x whenever x is sufficiently large. Thus, for x sufficiently large we have q log x x q log x...
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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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