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How does the function F(s) = 12*exp(-2s) / [(s-4)(s-3)] behave at infinity?

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steve10

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The topic "Zeros of the funcion in Laplace domain" has been locked, but there is some confusion needs to be cleared. Of course, I am not obliged doing this, but I am doing it for fun.
Actually, I have made a request to Pisoiu to unlocked that topic by apologizing the use of emotional words and promising no more similar words from me. There is no immediate comment or response from him(her), though.

I just want to clear up one thing, how does the function F(s) = 12*exp(-2s) / [(s-4)(s-3)] behave at infinity? I believe most of us are familiar with this, but I am also sure some don't know. Here is the conclusion:

Infinity is an essential singular point of F(s). In Complex Analysis, if the function has more than one different limiting values when the variable s approches one point (including infinity) s0 from different directions, then s0 is called the essential singular point of function F(s). In our example, infinity is such a point of F(s)=exp(-2s) / [(s-4)(s-3)], which can be easily seen. When s -> infinity from positive real axis, F(s) ->0, while if s->infinity from the negative real axis, then F(s) -> +infinity. So, infinity is an essential singular point of F(s). That means, among other things, infinity is not the zero of F(s).
 

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