How to apply squeeze theorem to this limit. I'm trying to solve $$\int_0^∞ e^{-x} \cos(x)\,dx$$
It is not hard to find that $$\int e^{-x} \cos(x)=\frac{1}{2}(e^{-x} \sin(x)-e^{-x} \cos(x))+C$$
From all this follows that
$$\lim_{t\to\infty}\int_0^te^{-x} \cos(x) \, dx = \frac{1}{2}\lim_{t\to\infty}(e^{-t} \sin(t) - e^{-t} \cos(t))+\frac{1}{2}$$
Notice that I have simplified already a lot the expression we are taking the limit of.
I have not been able to find this limit; a collegue student told me that I had to use the squeeze theorem, but I do not find how nor where. Any guides on how the theorem can help with this limit?

\begin{align}
& \lim_{t\to\infty} \left( e^{-t} \sin t - e^{-t} \cos t \right) \\[8pt]
= {} & \left( \lim_{t\to\infty} e^{-t}\sin t \right)  - \left( \lim_{t\to\infty} e^{-t} \cos t \right) \\
& \text{provided these last two limits exist} \\
& \text{(and are finite).}
\end{align}
Then, since $-1 \le \sin t \le 1,$ we have
$$
-e^{-t} \le e^{-t}\sin t \le e^{-t}.
$$
The first and third expressions above both approach $0$ as $t\to\infty.$
(And the one involving the cosine is done the same way.)
