Under what circumstances do $O(n^{a + \epsilon})$ algorithms imply $O(n^{a+o(1)})$ algorithms? Suppose that, for each $\epsilon &gt; 0$, there is an Turing machine $M_{\epsilon}$ that decides a language $L$ in time $O(n^{a + \epsilon})$. Is there a single algorithm deciding $L$ in time $O(n^{a + o(1)})$? (Here, the $o(1)$ term is measured in terms of $n$, the input length.) 

Does it make a difference if the algorithms $M_{\epsilon}$ are computable, or efficiently computable, in terms of $\epsilon$?

Motivation: in many proofs, it is easier to construct algorithms of time $O(n^{a + \epsilon})$ than the limiting algorithm $O(n^{a + o(1)})$. In particular, you need to bound the constant term in $O(n^{a + \epsilon})$ to pass to the limit $O(n^{a+o(1)})$. It would be nice if there is some general result you can invoke to pass to the limit directly.
The question is similar to the questions about constructive existence of the limit of a sequence of (constructive) objects. Usually if you can uniformly construct those objects (here $M\epsilon$) efficiently enough then you can show the existence of the limit constructively. 

For example, assume that we have a TM $N(k,x)$ which runs $M_{|k|^{-1}}$ on $x$ and its running time is $O(n^{a+|k|^{-1}})+O(|k|)$ (here the bounds are also uniform, e.g. something like $O(2^k.n^{a+|k|^{-1}})$ would not work). Then we can combine this uniform simulator with the function $(k,x)\mapsto x$ to obtain the machine $N(x,x)$ that runs in time $O(n^{a+o(1)})$. 

PS: $O(n^{a+o(1)})$ is a little bit ambiguous because of nesting of asymptotic notations, I am interpreting it as $n^{a+o(1)}$. Also we need $a$ to be not too small, e.g. at least $1$.