Laplace transform of the function $f_\epsilon (x)$

Question

Let $f_\epsilon (x)$ be defined as $$f_\epsilon (x)= \begin{cases} 1/\epsilon, & 0\leq x\leq \epsilon \\ 0, & x> \epsilon \end{cases}$$

I calculated the laplace transform of $f_\epsilon (x)$ to be $$L(f_\epsilon (x))=\frac{1-e^{-p\epsilon}}{p\epsilon}$$

I want to find the limit of this Laplace transform $\lim_\limits{\epsilon \to 0}L(f_\epsilon(x))$. However I'm not sure how to do this. Any help?

Answer 1

You can do a Taylor expansion of the exponential $e^{-p\epsilon}=1-p\epsilon+(p\epsilon)^2/2+...$. The $1$ will cancel, and what you are left with is $$\lim_{\epsilon\to0}\frac{p\epsilon+(p\epsilon)^2/2+...}{p\epsilon}=1$$

Answer 2

Apply L'Hospital rule $$\lim _{\epsilon \to 0} \frac{1-e^{-p\epsilon}}{p\epsilon}= \lim_{\epsilon \to 0} e^{-p\epsilon} =1$$