1873_solutions

1 we assume that f is a convex function on an

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Unformatted text preview: uÝ lim exp u 100 exp u 2 follows from the composition theorem for limits. We now discuss the limit lim sin x x /2 tan x  lim exp tan x log sin x x /2 We look first at the limit 249 lim tan x log sin x  lim x /2 x /2  log sin x cos x sin x lim sin x log sin x cos x lim x /2 x /2 log sin x cos x  lim x /2 Applying L’Hôpital’s rule to the latter limit we obtain log sin x lim  lim cos x x /2 x /2 cot x 0 sin x and so it follows from the composition theorem for limits that lim sin x tan x  exp 0  1 x /2 Now we discuss the limit lim x xÝ log x x . Since this limit can be expressed as log x x lim exp xÝ 2 and since log x x lim xÝ 2 0 we have lim x xÝ log x x  exp 0  1. Now we discuss the limit lim xÝ x log x x  1 log x1  lim xÝ log x exp 2 log x  1 exp  lim exp xÝ log x 2 log x  1 2 2 We look first at lim log x  1 xÝ 2 log x 2 It is possible to find this limit using L’Hôpital’s rule by expressing it as lim log x  1 xÝ log x log x  1  log x  lim xÝ log 1  1 x 1 log x 2 x which gives us lim xÝ x x 1 1 x2 1 log x 2 x 2 2 x 1 x 2 x which can be seen easily to be 0. Almost every student will want to do the problem this way. On the other hand, one can also use the mean value theorem to choose a number that we shall call f x between x and x  1 such that log x  1 2 log x 2 2 log f x log x  1 2 log x 2   . x1 x fx Since the function g defined by the equation log u gu  u for u  e, having a negative derivative, must be decreasing we have 250 1 2 log f x fx log x x log x  1 x1 1 and it follows that lim xÝ 2 log f x fx  0. From either method we deduce that lim xÝ x log x x  1 log x1 2. Given that   0, evaluate the limit lim x xÝ  exp 0  1. 2x  2  2x  1 2x  2   . Solution: Although L’Hôpital’s rule can be used to solve this problem, it is not the best approach. Instead we should define f x  x for all x  0 and, for a given positive number x, apply the mean value theorem to the function f on the interval 2x  1, 2x  2 to choose a number c between 2x  1 and 2x  2 such that f 2x  2 f 2x  1  f c  c  1 2x  2  2x  1   2x  2 2x  1 In the event that  1 we know from the fact that 2x  1  c  2x  2 that  2x  1  1 c  1  2x  2  1 and so x 2x  1  1 2x  2  2x  1  x 2x  2  1 x   2x  2 2x  2 2x  2  which we can express as  2x  1  x x 2x  1  1 x 2x  2 2x  2 2x  2 2x  2 2x  2  and we deduce from the sandwich theorem that 2x  2  2x  1   lim x xÝ 2 2x  2  In the event that 0    1 the inequalities are reversed but the result is the same. Try to handle this case yourself. 3. Evaluate the limits lim log x  1 xÝ lim xÝ log x  1 log x  log x  log x  1 log x x lim xÝ x log x For the latter two limits, check the limit value with Scientific Notebook. The limit lim log x  1 xÝ  lim log x  1 xÝ 2 log x  is an extension of the limit log x 2 that is discussed above. We understand  to be a given constant in this exercise. In the...
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