Prove that $\lim \limits_{T\to \infty} \frac{1}{T} \int_{-T/2}^{T/2} \cos(\omega t + \theta)dt = 0$ How to prove that $L=0$ if
$$L = \lim \limits_{T\to \infty} \frac{1}{T}\int\limits_{-T/2}^{T/2} \cos(\omega t + \theta)\, dt.$$
I want to say the numerator is equal to zero as $T$ approaches infinity because integrating a period function over an infinite range is zero, however, the denominator also grows to infinity.
Is that enough to prove that $L=0$?
 A: $$\left|\int_{-T/2}^{T/2} \cos( \omega t + \theta) \, dt\right|\le \frac{2}{|\omega|}=M$$
So $$|L| \le \lim_{T\rightarrow \infty} \frac{M}{T} = 0.$$
A: The case $\omega=0$ has already been addressed in comments. Now, assuming $\omega\neq0$, we have
$$\begin{align}
\int_{-T/2}^{T/2}\cos(\omega t + \theta)\,dt &= \left.\frac{1}{\omega}\sin(\omega t+\theta)\right|_{t=-T/2}^{t=T/2}\\
&=\frac{1}{\omega}\left(\sin\left(\omega \frac{T}{2}+\theta\right)-\sin\left(-\omega \frac{T}{2}+\theta\right)\right)\\
&=\frac{1}{\omega}\left(\sin\left(\frac{\omega T}{2}+\theta\right)+\sin\left(\frac{\omega T}{2}-\theta\right)\right)\\
&=\frac{2}{\omega}\sin\left(\frac{\omega T}{2}\right)\cos\left(\theta\right)\\
\end{align}$$
where we used the fact that $\sin(a+b)+\sin(a-b)=2\sin(a)\cos(b)$. Therefore, we have
$$L=\lim_{T\to\infty}\frac{2}{\omega T}\sin\left(\frac{\omega T}{2}\right)\cos\left(\theta\right)=\cos(\theta)\cdot\lim_{T\to\infty}\frac{2}{\omega T}\sin\left(\frac{\omega T}{2}\right)$$
If $\omega$ is not a function of $T$, then it behaves as a constant for the limit above and we can rewrite $x=\frac{\omega T}{2}$, which I assume is the context of OP's question. Then, $x\to\infty$ when $T\to\infty$ and
$$L=\cos(\theta)\cdot\lim_{T\to\infty}\frac{2}{\omega T}\sin\left(\frac{\omega T}{2}\right)=\cos(\theta)\cdot\lim_{x\to\infty}\frac{\sin(x)}{x}=0.$$
If the latter assumption is wrong and $w=f(T)$, the limit depends on the relationship between $\omega$ and $T$. As an example,
$$\begin{align}
\omega=\alpha/T&\Rightarrow L = \frac{2}{\alpha}\sin\left(\frac{\alpha}{2}\right)\cos(\theta).\\
\end{align}
$$
