If there is $x(t) = rect(\frac{t}{2})$, then its L.T will be $X(s) = 1/s(e^s - e^{-s})$. right? and after that i tried to draw them on S-Plane to check if poles exist. In the L.T result, it looks like having pole at $s=0$. but when $s\to 0$, numerator goes to zero too. so I concluded 'there is no pole'.
today i asked friend about 'how to check if there is poles or not using lhopital's rule?'. then he applied lhopital's rule on L.T result above.
$X(0) = 1/s(e^s - e^{-s})\ at\ s=0 \to e^s + e^s\ at\ s=0 = 2$.
he said the result of '2' is reason why there actually no pole of L.T result.
Q. I don't know why he apply lhopital's rule to checking existence of Poles.
can you tell me why?
Suppose $f(z) = g(z)/h(z)$ where $g(z)$ and $h(z)$ are analytic in a neighbourhood of $z=a$, and both $g(a) = h(a) = 0$ but $h$ is not identically $0$. Thus $a$ is a zero of $h$, let's say of multiplicity $m$, and of $g$, say of multiplicity $n$. Thus $a$ is an isolated singularity of $f$. Then if $n \ge m$, $a$ is a removable singularity of $f$ (thus not a pole), while if $n < m$ it is a pole of order $m-n$. Basically what your friend did was to show that $m=n=1$ in your example ($m=1$ because $h'(0) \ne 0$ and $n=1$ because $g'(0) \ne 0$).