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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
1x 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
1x
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 1x 1/x lim x 0
x0 and L’Hôpital’s rule rule guarantees us the equality
lim 1x
x e x0 1/x lim x
1 x 0 log 1 x log 1x
x exp x2 x0 as long as the latter limit exists. Now
lim
x0 0 x
1 x log 1 x
x2 exp log 1x
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
1x 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 x1
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.
 Fall '08
 STAFF
 Math, Calculus

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