The biggest thing that you’re doing wrong is trying to treat $\infty$ as if it were a number with which you can do arithmetic: it isn’t. You really do have to work with limits. Let’s start with the integral for some fixed value of $T$:
$\begin{align*} \int_{-T/2}^{T/2}e^{-2at}dt&=-\frac1{2a}\left[e^{-2at}\right]_{-T/2}^{T/2}=-\frac1{2a}\left(e^{-aT}-e^{aT}\right)\\ &=-\frac1{2a}\left(\frac1{e^{aT}}-e^{aT}\right)\\ &=\frac1{2a}\left(e^{aT}-\frac1{e^{aT}}\right)\\ &=\frac{e^{2aT}-1}{2ae^{aT}}\;. \end{align*}$
Thus,
$\begin{align*} \lim_{T\to\infty}\frac{1}T\int_{-T/2}^{T/2}e^{-2at}dt&=\lim_{T\to\infty}\frac{e^{2aT}-1}{2aTe^{aT}}=\lim_{T\to\infty}\left(\frac{e^{2aT}}{2aTe^{aT}}-\frac1{2aTe^{aT}}\right)\\ &=\lim_{T\to\infty}\frac{e^{2aT}}{2aTe^{aT}}-\lim_{T\to\infty}\frac1{2aTe^{aT}}\;. \end{align*}$
You should have no trouble evaluating $\lim_{T\to\infty}\dfrac1{2aTe^{aT}}$, and
$\lim_{T\to\infty}\frac{e^{2aT}}{2aTe^{aT}}=\lim_{T\to\infty}\frac{e^{aT}}{2aT}\;,$
which should also cause no trouble.