In doing some Laplace stuffs, I had to solve this limit somewhere in my calculation to proceed;
$$\lim_{T \rightarrow \infty} \arctan\left(\frac{T}{\omega}\right)$$
where the original function was defined as $f(t) = \ln\left(1 + \frac{\omega^2}{s^2}\right)$. Ordinarily, I know that the $\lim_{T \rightarrow \infty} \arctan\left(T\right) = \frac{\pi}{2}$. So I reasoned out that as $T \rightarrow \infty$, the expression $\frac{T}{\omega} \approx T$ and so the limit could be $\frac{\pi}{2}$.
- Please verify this conclusion for me.
- And indicate how this could be solved generally for inverse functions likes arctan.
Thanks.