If we consider the function $f(t)=e^{-at}\cos(t) v(t)$ (where $v(t)=1\; \forall t\geq 0$, and $0$ otherwise), then I have calculated its Fourier transform, which is
$$F(f)(\omega)=\frac{a+\omega i}{1+(a+\omega i)^2}$$
Can I take limits $a\to 0$ here to deduce the Fourier transform of $\cos(t)v(t)$?
Yes, this is a legitimate approximation argument. The Fourier transform is a continuous operator on the space of tempered distributions. The functions $f_a(t)=e^{-at}\cos(t) v(t)$ converge to $f(t)=\cos(t) v(t)$ in the sense of tempered distributions, meaning $\int f_a \varphi \to \int f \varphi$ for every Schwarz test function $\varphi$ (this holds by the dominated convergence, $|f\varphi|$ being the dominating function).
Therefore, the Fourier transforms of $f_a$ converge to the transform of $f$ in the sense of distribution. Specifically, $\hat f$ is a distribution represented by the function $\omega i/(1-\omega^2)$ in the sense of principal value, plus the delta functions pointed out in the comments below.