1873_solutions

# Interval hint use the preceding exercise this argument

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Unformatted text preview: x0 sin x x3 provided that the latter limit exists. Since lim 1 x0  lim 1 x0 cos x 3x 2 cos x  lim 3x 2  0 x0 L’Hôpital’s rule guarantees that lim 1 cos x  lim sin x 6x x0 3x 2 provided that the latter limit exists. You may know already that the latter limit is obtain it by using L’Hôpital’s rule one more time. x0 1 6 or you may To find the limit lim tan x x sin x x0 x we observe that lim tan x x0 x  lim x x0 sin x  0 and so L’Hôpital’s rule guarantees that 2 lim tan x x  lim sec x 1 sin x 1 cos x x0 x x0 provided that the latter limit exists. Although it is possible to use L’Hôpital’s rule on this latest expresssion, it is better to simplify instead. 247 2 lim sec x 1 1 cos x x0 1  lim x0 1  lim 1 1 x0 cos 2 x cos x cos 2 x cos x 1  cos x cos x cos 2 x  lim 1  cos x x0 cos 2 x 2 To find the limit lim 1  x 1/x x0 we make the observation that if x  1 and x 1x 1/x 0 we have log 1  x  exp x Since log 1  x x x0 1  lim exp lim log 1  x x we have lim 1  x 1/x x0 x0 and from the composition theorem for limits we deduce that log 1  x lim 1  x 1/x  lim exp x x0 x0  exp 1  e. Solution: To find the limit 1x x e lim x0 1/x we make use of the fact that whenever 1  x  0 we have log 1  x 1  x 1/x  exp x . It is easy to see that lim x0 log 1  x x 1 and so we see that lim e x0 1x 1/x  lim x  0 x0 and L’Hôpital’s rule rule guarantees us the equality lim 1x x e x0 1/x  lim x 1 x 0 log 1  x log 1x x exp x2 x0 as long as the latter limit exists. Now lim x0 0 x 1 x log 1  x x2 exp log 1x x  lim x0 1  x log 1  x 1  x x2 x exp lim 1  x log 1  x 1  x x2 x  e lim 1  x log 1  x x2 x  x0 x0 248 log 1  x x lim exp x0 1 1x log 1  x x Since lim 1  x log 1  x x0 x  lim x 2  0, x0 L’Hôpital’s rule guarantees us the equality 1  x log 1  x x 1  log 1  x 1 e lim  e lim 2x x0 x0 x2 as long as the latter limit exists. One more application of L’Hôpital’s rule (or a direct method) shows that 1  log 1  x 1 1 lim 2 2x x0 and we conclude that e 1  x 1/x lim e x 2 x0 x 100 exp log x lim xÝ lim x xÝ lim lim sin x 2 log x x ex x0 tan x x /2 lim xÝ log 1  x x2 x  2 1 lim xÝ x log x x  1 log x1 4 5x 3  8x 2 x4 2x  1 x We shall now discuss the limit x 100 exp log x lim xÝ 2 . It is possible to use L’Hôpital’s rule to yield the equation lim xÝ x 100 exp log x  lim xÝ 2 100x 99 exp log x 2 2 log x x but this fact doesn’t seem to be particularly useful. On the other hand, if we make the substitution u  log x then the limit we are trying to find takes on the form lim uÝ exp 100u exp u 2  u Ý exp 100u lim u2 and so the limit is obviously 0. The only question that remains is whether the idea of substitution u  log x really makes sense and whether the procedure is actually valid. What we are really saying, when making the substitution is that the desired limit can be expressed as exp log x exp log x lim xÝ 100 2 and that the equation lim xÝ exp log x exp log x 100 2...
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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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