# find the limit of this Laplace transform lim_(epsilon -> 0) L(f_(epsilon)(x)).

Let ${f}_{ϵ}\left(x\right)$ be defined as
${f}_{ϵ}\left(x\right)=\left\{\begin{array}{ll}1/ϵ,& 0\le x\le ϵ\\ 0,& x>ϵ\end{array}$
I calculated the laplace transform of ${f}_{ϵ}\left(x\right)$ to be
$L\left({f}_{ϵ}\left(x\right)\right)=\frac{1-{e}^{-pϵ}}{pϵ}$
I want to find the limit of this Laplace transform $\underset{ϵ\to 0}{lim}L\left({f}_{ϵ}\left(x\right)\right)$
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cesantedz
You can do a Taylor expansion of the exponential ${e}^{-pϵ}=1-pϵ+\left(pϵ{\right)}^{2}/2+...$ The 1 will cancel, and what you are left with is
$\underset{ϵ\to 0}{lim}\frac{pϵ+\left(pϵ{\right)}^{2}/2+...}{pϵ}=1$
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Oscar Burton
Apply L'Hospital rule
$\underset{ϵ\to 0}{lim}\frac{1-{e}^{-pϵ}}{pϵ}=\underset{ϵ\to 0}{lim}{e}^{-pϵ}=1$