Let $\mathcal{F}$ be a subset of $L^1(\Bbb{R})$, and suppose it has the property that for all $f\in \mathcal{F}$, the limit you write holds. Then, the closure $\overline{\mathcal{F}}$ of $\mathcal{F}$ in $L^1(\Bbb{R})$ has the same property as well, this is because for any $\phi\in \overline{\mathcal{F}}$ and any $f\in \mathcal{F}$ and any $a\in\Bbb{R}$, we have
\begin{align}
\left|\int_{\Bbb{R}}\phi(x)\sin(ax)\,dx\right|&\leq \left|\int_{\Bbb{R}}\phi(x)\sin(ax)\,dx-\int_{\Bbb{R}}f(x)\sin(ax)\,dx\right|+\left|\int_{\Bbb{R}}f(x)\sin(ax)\,dx\right|\\
&\leq\|\phi-f\|_{L^1(\Bbb{R})}+\left|\int_{\Bbb{R}}f(x)\sin(ax)\right|.
\end{align}
First, take the $\limsup$ of both sides as $a\to\infty$ (we take $\limsup$ since $\lim$ might not exist) to get
\begin{align}
\limsup_{a\to\infty}\left|\int_{\Bbb{R}}\phi(x)\sin(ax)\,dx\right|&\leq \|\phi-f\|_{L^1(\Bbb{R})}+0.
\end{align}
Now, since $f\in\mathcal{F}$ was arbitrary, by taking a sequence $f_n\in \mathcal{F}$ which converges to $\phi$ in $L^1(\Bbb{R})$, the RHS vanishes, and so we get
\begin{align}
\limsup_{a\to\infty}\left|\int_{\Bbb{R}}\phi(x)\sin(ax)\,dx\right|&\leq 0.
\end{align}
The $\limsup$ is clearly also $\geq 0$, so it is equal to $0$. If the $\limsup$ of a non-negative quantity vanishes, then the limit exists and is also equal to $0$. Hence, we conclude that every function in $\overline{\mathcal{F}}$ has the quoted property. If you don’t like $\limsup$, then feel free to transcribe everything above into an $\epsilon$-$\delta$ type proof.
One more small observation we can make is that if a collection of functions has this limit property, then the linear span (i.e finite linear combinations) also has this property.
With this general observation in mind, we see that in order to prove the limit in question holds for all $L^1$ functions, it suffices to fix a subset $\mathcal{F}$, whose linear span is dense, and prove it for these functions. Some simple examples of such sets, for which proving the statement becomes easy are:
- $\mathcal{F}$ being the set of indicator functions of intervals, i.e the set of all $\chi_{[\alpha,\beta]}$ with $\alpha,\beta\in\Bbb{R}$. Here, you can directly integrate and calculate the limit.
- $\mathcal{F}$ being the subspace of Schwartz functions, or smooth compactly supported functions or $C^1_c$. Here, using integration by parts once, you can easily prove the limit is $0$.
There are some other tricks as well from proving the Riemann-Lebesgue lemma, but this idea of proving things for a dense class of objects, and then taking limits is a very common theme in analysis.
