# How do you find the limit of sec3xcos5x as x approaches infinity?

Mar 29, 2016

There is no limit. (The limit does not exist.)

#### Explanation:

Let $f \left(x\right) = \sec \left(3 x\right) \cos \left(5 x\right) = \cos \frac{5 x}{\cos} \left(3 x\right)$.

$f \left(x\right) = 1$ whenever $x$ is an integer multiple of $\pi$. To show this we can look at cases.
If $x = k \pi$ for even $k$, then $\cos \left(5 x\right) = \cos \left(3 x\right) = 1$
If $x = k \pi$ for odd $k$, then $\cos \left(5 x\right) = \cos \left(3 x\right) = - 1$
In either case, we have $f \left(x\right) = 1$.
Of course this occurs infinitely many times as $x \rightarrow \infty$.

Every time $x$ is an odd multiple of $\frac{\pi}{2}$ we have one of three cases. One is enough to establish the nonexistence of a limit of $f \left(x\right)$ as $x \rightarrow \infty$.

If $\cos \left(5 x\right) = 0$ and $\cos \left(3 x\right) \ne 0$, then $f \left(x\right) = 0$
This also occurs infinitely many times as $x \rightarrow \infty$.

As $x \rightarrow \infty$, $f \left(x\right)$ hits both $0$ and $1$ infinitely many times. So, we see that $f \left(x\right)$ cannot approach a limit.