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




Re: [apcalculus] Limit problem
Posted:
Sep 10, 2012 7:52 PM


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 jeffrey.stuart@plu.edu
==== Course related websites: http://apcentral.collegeboard.com/calculusab http://apcentral.collegeboard.com/calculusbc  To search the list archives for previous posts go to http://lyris.collegeboard.com/read/?forum=apcalculus



