How is it true that for large $t$, $(1+O(1/t))e^{-2\ln t O(1/t)}=1+O(\ln t/t)$?

Question

The title pretty much says it all. At some point in large time analysis, the following claim popped out but I don't see how it is true: For sufficiently large $t>0$,

$$ \frac{2\ln t}{t}(1+O(1/t))e^{-2(\ln t) O(1/t)}=\frac{2\ln t}{t}(1+O((\ln t)/t)) $$

It'd be nice if someone could explain what's going on here

Answer 1

Hint. One may just write, as $t \to \infty$, $$ \begin{align} (1+O(1/t))e^{-2\ln t \, O(1/t)}&=(1+O(1/t))e^{ O((\ln t)/t)} \\\\&=(1+O(1/t))(1+O((\ln t)/t)) \\\\&=1+O((\ln t)/t)+O(1/t)+O((\ln t)/t^2) \\\\&=1+O((\ln t)/t) \end{align} $$ as announced.