Big O equality

[Ozma]

My textbook says that for $u \to \infty$ it is $u-1+2\exp[-2u+2+O(e^{-2u})]+O(e^{-4u})=u-1+2\exp(-2u+2)+O(e^{-4u})$. I don't understand why this equality holds, it seems like that the term $2\exp[O(e^{-2u})]$ was substituted with $1$. I know that $f(x)=O(g(x))$ as $x \to x_0$ if there exists a constant $K$ such that $|f(x)|\le K|g(x)$ in a neighborhood of $x_0$, but I don't understand how the definition can be applied in the equality I have written: I tried to work on the fact that $O(e^{-2u})$ means a function that is bounded by $Ke^{-2u}$, hence the exponential $2e^{O(e^{-2u})}$ can be estimated by $2e^{Ke^{-2u}}$, which is bounded as well. at least when $u \to \infty$. But this would give me that $2e^{O(e^{-2u})}=O(1)$, not $1$. Can someone explain me this, please?

 

[blamocur]

I am guessing their claim is that $-2u+2+O(e^{-2u}) = -2u +2$ because $e^{-2u}$ is infinitely smaller than $u$ when $u\rightarrow\infty$