Question

How do I find the limit as x approaches infinity of a trigonometric function?

`lim_(x->oo)(x^2csc3xtan6x)/(cos7xcot^2x)`

Answer 1

The limit does not exist...

Explanation:

First consider:

`f(x) = (csc 3x tan 6x)/(cos 7x cot^2 x)`

The various constituent trigonometric functions have periods:

`(2pi)/3, pi/6, (2pi)/7, pi`

The least common multiple of these is `2pi`. Hence `f(x)` has period `2pi` or a factor thereof. In fact we can tell that it has period exactly `2pi` since `7` is prime.

When `x=pi/14` we find that `cos 7x = cos (pi/2) = 0` and all of the other trigonometric functions are non-zero.

If we take a small interval around `x = pi/14` then `cos 7x` changes sign while the other trigonometric functions retain their signs. Hence `f(x)` changes sign in a small interval around `x = pi/14`.

Since `cos 7x` is in the denominator, this means that `f(x)` has a vertical asymptote with different signs on either side of the asymptote at `x=pi/14 + 2npi` for any integer `n`.

Next note that `x^2->oo` as `x->oo` (or `x->-oo`).

Note also that all the trigonometric functions are continuous on their various domains.

Hence:

`(x^2 csc 3x tan 6x)/(cos 7x cot^2 x)`

is unbounded and takes every value in `(-oo, oo)` repeatedly as `x->oo` (or `x->-oo`). It definitely does not converge to a limit.