#### Answer

$\lim_{t\to0}g(t)/h(t)$ can still exist and does not have to equal $0$.

#### Work Step by Step

1) Even though $g(0)=h(0)=0$, $$\lim_{t\to0}\frac{g(t)}{h(t)}$$ can still exist.
Take this example: $g(t)=t^3$ and $h(t)=t$
Here both $g(t)$ and $h(t)$ are defined for all values of $t$ and $g(0)=h(0)=0$. However, $$\lim_{t\to0}\frac{g(t)}{h(t)}=\lim_{t\to0}\frac{t^3}{t}=\lim_{t\to0}t^2=0^2=0$$
So $\lim_{t\to0}g(t)/h(t)$ still exists.
2) Again, $\lim_{t\to0}g(t)/h(t)$ also does not have to be $0$.
Another example: $g(t)=t^2-t$ and $h(t)=t$
Both $g(t)$ and $h(t)$ are defined for all values of $t$ and $g(0)=h(0)=0$. However, $$\lim_{t\to0}\frac{g(t)}{h(t)}=\lim_{t\to0}\frac{t^2-t}{t}=\lim_{t\to0}(t-1)=0-1=-1$$
So $\lim_{t\to0}g(t)/h(t)$, in this case, does not equal $0$.