Simple asymptotic analysis problem I came across a problem that I tried to formalize as follows:
Let say i have two functions $x(t)$ and $y(t)$ such that for $t \rightarrow t_0$
$$
\left\{
\begin{array}
 \;y(t) \rightarrow -\infty \\
 x(t) \rightarrow -\infty \\
 y(t) = o(x(t))
\end{array}
\right.
$$ 
Shouldn't be in such case $ 2^{y(t)} = o(2^{x(t)})$ ? If yes I struggle to prove it is true... otherwise what can we say instead? What about if we don't assume $y(t),x(t) \rightarrow - \infty$?
PS. Just to be clear.. by $y(t) = o(x(t))$ I mean that $\frac{y(t)}{x(t)} = 0$, for $t \rightarrow t_0$.

No, $ 2^{y(t)} = O(2^{x(t)})$ is false.
Consider the example 
$$ \left\{ \begin{array}  y(t) = -\frac1{|t-t_0|} \\  x(t) = -\frac2{|t-t_0|} \\ y(t) = \frac12 x(t) = O(x(t)) \end{array} \right. $$
But if we let $u = \frac1{|t-t_0|}$, then 
$2^{-u}$ is not $O(2^{-2u})$, although it is $O(\sqrt{2^{-2u}})$:  as $u\to\infty$, the ratio of $2^{-u}$  to $2^{-2u}$ exceeds any constant $C$.
