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 firstname.lastname@example.org