Proof $\frac{\sin(ax)}{x} \rightarrow a$ as $x \rightarrow 0$ in the context of a Laplace Transformation I'm finding it difficult to get my head around 

$\frac{\sin(ax)}{x} \rightarrow a$ as $x \rightarrow 0,$

however for the Laplace transform of this function,

$\mathcal{L} \{ \frac{\sin(ax)}{x} \} = \arctan(\frac{a}{s}),$

surely, $\arctan(\frac{a}{s}) \rightarrow 0$ as $s \rightarrow \infty$? 

But to retain the behaviour seen in the prior limit, as

$\mathcal{L}^{-1}\{ \frac{a}{s} \} = a$,

we would require the nonsensical situation of

$\arctan(\frac{a}{s}) \rightarrow \frac{a}{s}$ as $s \rightarrow \infty.$ 

Does this create an issue for our description of this function as $t \rightarrow 0$?

Is there a relationship between $s$ and $a$ that prevents us from taking this limit? I feel like I am missing something fundamental here.
The final value theorem for Laplace's transform actually states:

$$
\lim_{s\to \infty} sF(s) = \lim_{t\to 0^+} f(t)
$$

Thus:
$$
\lim_{s\to \infty} s \left[\text{sgn}(a) \frac{\pi}{2}-\arctan \left( \frac{s}{a} \right) \right] = \lim_{t\to 0} \frac{\sin(at)}{t}
$$

Check the result for the Laplace transform with Wolfram Alpha.

The limit above for $s$ can be computed with L'Hopital's rule and yields $a$, but that I guess is possibly pointless when the limit for the sine is being put into question.
The final value theorem for Laplace's transform actually states:

$$
\lim_{s\to \infty} sF(s) = \lim_{t\to 0^+} f(t)
$$

Thus:
$$
\lim_{s\to \infty} s \left[\text{sgn}(a) \frac{\pi}{2}-\arctan \left( \frac{s}{a} \right) \right] = \lim_{t\to 0} \frac{\sin(at)}{t}
$$

Check the result for the Laplace transform with Wolfram Alpha.

The limit above for $s$ can be computed with L'Hopital's rule and yields $a$, but that I guess is possibly pointless when the limit for the sine is being put into question.