Does $∩_{ε>0} \mathrm{DTIME}(O(n^{2+ε})) = \mathrm{DTIME}(n^{2+o(1)})$?
Here is a counterexample, i.e. a language with an $O(n^{2+ε})$ algorithm (using multitape Turing machines) for every $ε&gt;0$, but not uniformly in $ε$:
Accept $0^k 1^m$ iff $k&gt;0$ and the $k$th Turing machine halts in less than $m^{2+1/k}$ steps on the empty input.  Other strings are rejected.

For every $ε$, we get an $O(n^{2+ε})$ algorithm by hardcoding all sufficiently small nonhalting machines, and simulating the rest.

Now, consider a Turing machine $M$ deciding the language.

Let $M'$ (on the empty input) be an efficient implementation of the following:
for $n$ in 1,2,4,8,...:
&nbsp;&nbsp;&nbsp;&nbsp; use $M$ to decide whether $M'$ halts in $&lt;n^{2+1/M'}$ steps.
&nbsp;&nbsp;&nbsp;&nbsp; halt iff $M$ says that we do not halt but we can still halt in $&lt;n^{2+1/M'}$ steps.

By correctness of $M$, $M'$ does not halt, but $M$ takes $Ω(n^{2+1/M'})$-steps on input $0^{M'} 1^{n-M'}$ for infinitely many $n$.  (If $M$ is too fast, then $M'$ would contradict $M$.  The $Ω(n^{2+1/M'})$ bound depends on $M'$ simulating $M$ in linear time and otherwise being efficient.)