How do I evaluate $\lim_{h \to \infty} e^{h(1-s)}$? 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?
 A: There's a couple approaches to computing the Laplace transform of $e^t$ but the "slickest" way is just to use the derivative rule:
$$\mathcal{L}_t[f'(t)](s) = s\mathcal{L}_t[f(t)](s) - f(0).$$
Then you can notice that the derivative of $e^t$ is just $e^t$, so you can solve for $\mathcal{L}_t[e^t]$ in the above equation. This gives $\mathcal{L}_t[e^t] = \frac{1}{s-1}$.
The derivative rule is just computed with integration by parts, which you could use for $e^t$ directly. See this page.

If you look at the page I linked, it sates:

The Laplace transform existence theorem states that, if $f(t)$ is
  piecewise continuous on every finite interval in $[0,\infty)$
  satisfying $$f(t) \leq Me^{at}$$ for all $t\in[0,\infty)$, then
  $\mathcal{L}_t[f(t)](s)$ exists for all $s > a$.

So basically, given $a$ you need to find an $M$ such that $\log M \geq (1-a)t$ for every $t\in[0,\infty)$. You can obviously do this when $a \geq 1$ because the RHS is always negative or $0$ in that case. If $a < 1$ you can't, because the RHS is unbounded. So $a = 1$ is the best choice here and $\mathcal{L}_t[f(t)](s)$ exists for $s > 1$.

