I'm not exactly sure how to approach this question.
My (rough) work so far:
Since we can assume $\lim_{x \to 0} f(x) = L$, we can conclude that $\forall \varepsilon > 0$, even $\frac{1}{3} \varepsilon$, $\exists \delta > 0$ s.t. $0 < \lvert 2x \rvert < \delta\ \Rightarrow\ \lvert f(2x) - L \rvert < \frac{1}{3} \varepsilon$.
Since $\vert f(2x) - L \rvert < \frac{1}{3} \varepsilon$, it follows that $\lvert 3f(2x) - 3L \rvert < \varepsilon$, which implies $\lim_{x \to 0} \left[ 3f(2x) \right] = 3L$, as required.
I don't quite understand how this should be done; any help is appreciated.
We need to show that
$$\forall \epsilon>0\quad\exists \delta>0\quad \forall x\quad |x|<\delta \quad|3f(2x)-3L|=3|f(2x)-L|<\epsilon \iff |f(2x)-L|<\frac{\epsilon}3$$
then since $\lim_{x \to 0} f(x) = L$
$$\forall \epsilon_1>0 \quad\exists \delta_1>0\quad \forall x\quad |x|<\delta_1 \quad |f(x)-L|<\epsilon_1 $$
and it suffices to take $\epsilon_1=\frac{\epsilon}3$ and $\delta=\frac{\delta_1}{2}$.