Thomas Calculus 13th [Solutions]

Section 4.5 Indeterminate Forms and LHôpitals Rule 289
89. (a) We should assign the value 1 to
f ( x) (sin x ) x to make it continuous at x 0.
(b)
ln(sin x) ln(sin x)
(cos x)
2
ln f ( x) x ln(sin x) lim ln f ( x) lim lim lim x
0
tan x
x 0 x 0 x 0
x
1 1
x
x
0
lim 2x
0 lim f ( x) e 1
2
x 0 sec x x 0
(c) The maximum value of f ( x ) is close to 1 near the point x 1.55 (see the graph in part (a)).
(d) The root in question is near 1.57.
1
sin x
1
x
2
90. (a) When sin x 0 there are gaps in the sketch.
The width of each gap is .
tan x
(b) Let f ( x) (sin x)
ln f ( x) (tan x) ln(sin x) lim ln f ( x)
x
ln(sin x)
lim
cot x
x
lim
x
2 2
2 2
1
sin x
2
2
(cos x)
csc
lim cos x
0
( csc )
0 lim f ( x
x
) e 1.
x
x
0
Similarly, lim f ( x) e 1. Therefore,
x
lim f ( x) 1.
x
2
2
x
(c) From the graph in part (b) we have a minimum of about 0.665 at x 0.47 and the maximum is about
1.491 at x 2.66.
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