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Topic: [ap-calculus] Limit problem
Replies: 3   Last Post: Sep 11, 2012 5:39 AM

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Jeff Stuart

Posts: 1,086
Registered: 12/6/04
Re: [ap-calculus] Limit problem
Posted: Sep 10, 2012 7:52 PM
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Dear Colleagues:

Some of you have mistakenly read a limit problem that you expected
rather than the one that was asked.
Specifically, the limit is question is as x --> infinity rather than
as x --> 0. I would agree that

lim (x --> 0) tan(x)/x = 1 since the limits as x --> 0 of sin(x)/x
and 1/cos(x) are both 1.

However, the question asked about

lim (x --> oo ) tan(x)/x, which is both more interesting and less trivial.

First, note that if f(x) is bounded as x --> oo, then lim (x --> oo)
f(x)/x = 0. However, the function tan(x) is not bounded when x --> oo
, so this argument cannot be invoked.

By considering x = pi * k + pi/2 - epsilon and x = pi* k - pi/2 +
epsilon where epsilon is small and positive, and where k is a positive
integer and k -->oo, we can see that if there is a limit, it must be
zero since the first sequence produces positive values and the second
sequence produces negative values.

Pick a positive integer k. Then lim (x --> pi*k + pi/2 from
below) tan(x)/x = +oo . Thus there is a sequence of points that
diverges to oo at which tan(x)/x attains the "value" of +oo, which
means that as x --> oo, tan(x)/x is not bounded, and hence not convergent.

Professor Jeff Stuart, Chair
Department of Mathematics
Pacific Lutheran University
Tacoma, WA 98447 USA
(253) 535 - 7403

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