I'm messing around with Laplace, and was trying to find the transform of $e^{t}$ and I have to evaluate $$\lim_{h \to \infty} e^{h(1-s)}$$ I figure if $s=1$, the limit is $1$. If $0≤s<1$, the limit is $\infty$. If $s>1$, the limit is $0$.
WolframAlpha tells me the answer is complex infinity. I don't know what that is.
When I ask WolframAlpha for the transform of $e^t$, it says $\frac{1}{s-1}$, which would assume that s>1, though it doesn't mention this. What if $s$ isn't greater than 1?