I have to compute this limit:
Suppose that $|\alpha| \neq |\beta |$, compute the following limit: \begin{align} \lim_{T \rightarrow \infty} \frac{1}{T} \int_{0}^{T} \sin(\alpha x)\cos(\beta x) \, dx \end{align}
What I've done is this: \begin{align} & \int_0^T \sin(\alpha x)\cos(\beta x) \, dx \\[8pt] = {} & \frac{1}{2(\alpha-\beta)} + \frac{1}{2(\alpha+\beta)}-\frac{\cos((\alpha-\beta)T)}{2(\alpha-\beta)}-\frac{\cos((\alpha+\beta)T)}{2(\alpha+\beta)} \end{align}
Then,
\begin{align} \lim_{T \rightarrow \infty}\frac{1}{T}\int_0^T \sin(\alpha x) \cos(\beta x) \,dx=\lim_{T \rightarrow \infty}\frac{1}{T}\left [ \frac{1}{2(\alpha-\beta)}+\frac{1}{2(\alpha+\beta)}-\frac{\cos((\alpha-\beta)T)}{2(\alpha-\beta)}-\frac{\cos((\alpha+\beta)T)}{2(\alpha+\beta)} \right ] \end{align}
But I'm not sure how can I conclude. What I think that I can do is this:
\begin{align} \underbrace{\lim_{T \rightarrow \infty}\frac{1}{T}}_{=0}\cdot \underbrace{\lim_{T \rightarrow \infty}\left [ \frac{1}{2(\alpha-\beta)}+\frac{1}{2(\alpha+\beta)}-\frac{\cos((\alpha-\beta)T)}{2(\alpha-\beta)}-\frac{\cos((\alpha+\beta)T)}{2(\alpha+\beta)} \right ]}_{\text{I don't know how to compute it}} \end{align}
So intuitively I think this must be zero. How can I conclude? Is there any other way to solve it? I really appreciate your help.