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Reading the article on the Laplace Transform in Wolfram MathWorld, I found the proof that $\mathcal{L}[f'(t)] = sF(s) - f(0)$.

Laplace transform of a derivative

I understand the first and second steps, but I don't understand the third one. Why is it that $lim_{a \to \infty} [e^{-sa} f(a)] = 0$? $e^{-sa}$ does get closer to 0 when $-sa$ approaches to $\infty$, but why does $f(a)$ get closer to 0? As far as I know, $f(a)$ could be anything, so it could be possible that $lim_{x \to \infty} f(a)$ doesn't exist.

What guarantees that the limit of $f(a)$ always exists?

I hope I'm not missing some simple property of limits here.

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up vote 3 down vote accepted

Recall that the Laplace transform requires $f$ to be of exponential order, e.g. $f(x)=O(e^{cx})$ as $x\to\infty$ for some $c>0$.

This means that $\frac{f(x)}{e^{ax}}$ is bounded as $x\to\infty$. Re-interpreting this in terms of the above limit, we see it is indeed $0$ when $s$ is sufficiently large (this restricts the domain of $L\{f'\}(s)$ by the way) since by taking $s>c$ (whatever $c$ is for this $f$) we then have $f=o(e^{sa})$ which means $\frac{f(a)}{e^{as}}\to0$ as $a\to\infty$ whenever $s>c$. And the limit doesn't need to exist for $f$; indeed $\lim_{a\to\infty}f(a)$ is unlikely to exist (consider $\sin a$), but that's not what is important here.

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Oh, you're right. I forgot about the order of the function. Thanks. – user1002327 Oct 28 '12 at 2:09
I made a couple edit to help clarify the situation in more detail. – Taylor Martin Oct 28 '12 at 2:17

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