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