calculating the fourier transform of
$$f(t)=e^{-|t|}$$
yields
$$F(x)=\int_{-\infty}^{0}e^{(1-ix)t}\,dt + \int_{0}^{+\infty}e^{-(1+ix)t}\,dt = \left[ \frac{e^{(1-ix)t}}{(1-ix)} \right]_{-\infty}^{0} + \left[ \frac{e^{-(1+ix)t}}{-(1+ix)} \right]_{0}^{+\infty}.$$
i know that
$$\lim_{t \to -\infty} e^{t}=0,\quad \lim_{t \to 0} e^{t}=1,\quad \lim_{t \to +\infty} e^{t}=+\infty.$$
however, i am rather unfamiliar with limits of functions taking complex arguments. do the limits
$$\lim_{t \to -\infty}e^{it},\quad \lim_{t \to +\infty}e^{it}$$
exist? how can i proceed in the above expression?
The limit $$ \lim_{t \rightarrow \infty} \frac{e^{-(1 + ix)t}}{-(1 + ix)} $$ exists and is $0$ due to the decay of $e^{-(1+ix)t}$ as $t \rightarrow \infty$ (and similarly for the other endpoint). To see this, just note that if we take absolute values we have \begin{align*} \left | \frac{e^{-(1+ix)t}}{-(1 + ix)} \right | &= \frac{|e^{-(1+ix)t}|}{|1 + ix|} \\ &= \frac{e^{-t}}{\sqrt{1 + x^2}} \\ &\leq e^{-t}. \end{align*} Thus, by the squeeze theorem, we have \begin{align*} \lim_{t \rightarrow \infty} \left | \frac{e^{-(1+ix)t}}{-(1 + ix)} \right | = 0, \end{align*} which is equivalent to \begin{align*} \lim_{t \rightarrow \infty} \frac{e^{-(1+ix)t}}{-(1 + ix)} = 0. \end{align*}