We have the definition of what it means to say that $f(x)$ tends to $M \in \mathbb{R}$
It is: $\forall \epsilon > 0, \exists \space \delta > 0$ such that $x \in (a,b)$ and $0 < |x-c| <\delta \implies |f(x) - M| < \epsilon$.
Now, I understand that we want to show the case is true for $f(3x)$ but directly showing that from the definition is escaping me.
So far I have tried:
Assuming the result true for $y$ and then letting $y=3x$ but I'm not sure that's showing the result directly from the definition.
And I've considered using uniqueness of limits, but the question strictly states no results about limits other than the definition above can be used.
I know that $\sin(3x)$ is bounded between $-1$ and $1$ but I'm not sure how this will help.
In response to a comment:
We know: $0 < |x| < \delta' \implies |\frac{\sin{x}}{x} - 1| < \epsilon'$
and we want to show $0 < |x| < \delta \implies |\frac{\sin{3x}}{3x} - 1| < \epsilon$
We have $f(x) = \frac{\sin{x}}{x}$
We know, for some $\delta > 0$ $$0 < |x| < \delta \implies |f(x) - 1| < \epsilon$$
We want to show, $\exists \space \delta' > 0$ such that $$0 < |x| < \delta' \implies |f(3x) - 1| < \epsilon$$
If we let $\delta'= \frac{\delta}{3}$, we have $$0 < |x| < \frac{\delta}{3} \iff 0 < |3x| < \delta \implies |f(3x) - 1| < \epsilon.$$
And we are done.