I have found it hard to find an intuitive proof for the Fourier transform of a complex exponential, $f(t) = e^{i\omega_0t} $, via direct calculation. The vast majority of pages online use the inverse FT of $ 2\pi\delta(\omega-\omega_0)$. Looking online, I found this definition of the delta function, and have attempted to use it in a derivation. I would like to know if the following would be acceptable in a standard engineering signals course: $$\begin{align*} F(\omega) &= \lim_{T\to\infty} \int_{-T}^{T}e^{i\omega_0t}e^{-i{\omega}t} dt \\ &= \lim_{T\to\infty} \int_{-T}^{T}e^{-it(\omega-\omega_0)} dt \\ &= \lim_{T\to\infty} \frac{e^{iT(\omega-\omega_0)}-e^{-iT(\omega-\omega_0)}}{i(\omega-\omega_0)}\\ &=\lim_{T\to\infty} \frac{2{\sin}[T(\omega-\omega_0)]}{\omega-\omega_0} \end{align*}$$
Now, we can perform a change of variables $\tau = \frac{\pi}{T}$, causing our limit to now approach zero (From here on is what I am unsure of): $$\begin{align*} F(\omega) &=\lim_{\tau\to0} \frac{2\sin[\frac{\pi(\omega-\omega_0)}{\tau}]}{\omega-\omega_0} \\ &= 2\pi \lim_{\tau\to0}\frac{\sin[\frac{\pi(\omega-\omega_0)}{\tau}]}{\pi(\omega-\omega_0)} \\ \end{align*}$$
According to the Wikipedia article linked above: $$ \lim_{\tau\to0}\frac{\sin[\frac{\pi(\omega-\omega_0)}{\tau}]}{\pi(\omega-\omega_0)} = \delta(\omega-\omega_0) $$
And therefore we can conclude:
$$ F(\omega) = 2\pi\delta(\omega-\omega_0) $$
Overall, what do people think about this derivation?