Is this solution for the limit correct? We have the limit:

$\lim_{t \downarrow 0} t^{(t^a)}$ with $a \in \mathbb{R}$.

My solution was as follows:
We know that $\lim_{t \downarrow 0} t^a = 0^a = 0$ with $a \in \mathbb{R}$. So we have $\lim_{t \downarrow 0} t^{(t^a)} = \lim_{t \downarrow 0} t^0 = 1.$
Is this correct? Am I allowed to say that  $\lim_{t \downarrow 0} t^{(t^a)} = \lim_{t \downarrow 0} t^0$?
 A: The analysis, and answer, depends on the value of the constant $a$, so we consider various cases.
(i) Suppose that $a$ is negative. Then as $t$ approaches $0$ from the right, $t^a$ becomes very large. So $t^{t^a}$ approaches $0$, fast.
(ii) Suppose next that $a=0$. Then for $t$ positive, $t^a$ is identically $1$. Thus when $a=0$ our function, for positive $t$, is simply $t$. It follows that in the case $a=0$  the function $t^{t^a}$ has limit $0$ as $t$ approaches $0$ from the right.
(iii) Finally, let $a\gt 0$. I think the limit is less obvious in this case. Your limit-taking procedure, which involved taking the limit of part of the expression, is not justified, and can give the wrong answer. However, here it happened to give the right numerical answer. 
We can express $t^{t^a}$ as $\exp(t^a\ln t)$. Now find what happens to $t^a\ln t$ as $t\to 0^+$. Maybe it will be clearer if we let $w=1/t$. Then
$$t^a\ln t=-\frac{\ln w}{w^a}.$$
In the long run $\ln w$ grows more slowly than any positive power of $w$, so $\lim_{w\to\infty}-\frac{\ln w}{w^a}=0$.  This can be verified in various ways, including L'Hospital's Rule.
Finally, because $t^a\ln t\to 0$, we conclude that $t^{t^a}\to \exp(0)$.  So our limit is $1$.
