Let: $$ \lim_{t\to t_0}\phi(t) = a $$ where $\phi(t)\ne a$ and $t\ne t_0$ in the neighbourhood of $t_0$. Prove that: $$ \begin{align*} f(x) \stackrel{x\to x_0}{=} \mathcal{o}(g(x)) &\implies f(\phi(t)) \stackrel{t\to t_0}{=} \mathcal{o}(g(\phi(t)))\tag 1\\ f(x) \stackrel{x\to x_0}{=} \mathcal{O}(g(x)) &\implies f(\phi(t)) \stackrel{t\to t_0}{=} \mathcal{O}(g(\phi(t)))\tag 2 \end{align*} $$
I'm having difficulties with proving the above. I've started by writing down the definitions of what's given in the statement. First, we are given that: $$ \lim_{t\to t_0}\phi(t) = a \iff \forall\epsilon>0\ \exists \delta_\epsilon > 0: |t-t_0| < \delta_\epsilon \implies |\phi(t) - a| < \epsilon $$
Also $f(x)$ is asymptotically smaller than $g(x)$ in the neighbourhood of $x_0$, namely: $$ \lim_{x\to x_0} \frac{f(x)}{g(x)} = 0 \iff \forall\epsilon>0\ \exists \delta_\epsilon > 0: |x-x_0| < \delta_\epsilon \implies \left|\frac{f(x)}{g(x)}\right| < \epsilon $$
Using the above, I need to somehow arrive at the following: $$ \lim_{t\to t_0} \frac{f(\phi(t))}{g(\phi(t))} = 0 $$
The problem is I'm not sure where to start. Could someone please share some kind of sketch on how to proceed?