2. Compare the infinitesimals f(x) = tan² x and g(x) = tan(sin x), as x → 0.

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
Section: Chapter Questions
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Exercise 2 I have attached the solution as well. In the solution I don’t understand why they put the denominator on top and switched with with something else and then they went back to the original. And another thing I don’t get is how can I know that f is infinitesimal of higher order than g why not vice versa?
and g(x)=√x+7-√7 are infinitesimals
Exercise 1. Verify that
f(x)=√x+5-√5
of the same order, as x → 0; then, determine c ER so that g~ cf, as x 0.
Exercise 2. Compare the infinitesimals f(x) = tan² x and g(x) = tan(sin x), as x → 0.
Transcribed Image Text:and g(x)=√x+7-√7 are infinitesimals Exercise 1. Verify that f(x)=√x+5-√5 of the same order, as x → 0; then, determine c ER so that g~ cf, as x 0. Exercise 2. Compare the infinitesimals f(x) = tan² x and g(x) = tan(sin x), as x → 0.
Exercise 2. Let's compute the limit of the quotient f/g:
f(x)
X
lim
x+0 g(x)
tan²
lim
*→0 tan (sin x)
Being lim
x40
Thus,
tan (sin x)
sin x
-
lim
y-0
tan y
Y
-
1, it follows that tan (sin x) ~ sinx, as x → 0.
tan² x
tan² x
lim
lim
x0 tan(sin x) x 0 sin x
Therefore, f is an infinitesimal of higher order than g, as x → 0.
= 0.
Transcribed Image Text:Exercise 2. Let's compute the limit of the quotient f/g: f(x) X lim x+0 g(x) tan² lim *→0 tan (sin x) Being lim x40 Thus, tan (sin x) sin x - lim y-0 tan y Y - 1, it follows that tan (sin x) ~ sinx, as x → 0. tan² x tan² x lim lim x0 tan(sin x) x 0 sin x Therefore, f is an infinitesimal of higher order than g, as x → 0. = 0.
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