Question: Prove that $\sup f([0,\infty))=1$, for $f(x)=\cos x - e^{-x}$.
My attempt: I started by showing that $\forall x \in [0,\infty)$, $f(x) \le 1$ as follows:
$$f(x) = \cos x - e^{-x} = \cos x - \frac{1}{e^x} \le 1 - \frac{1}{e^x} \le 1.$$
Here I used the fact that $\cos x \le 1$ and $e^{x}$ is positive.
Now I want to show that $\forall \epsilon > 0$, $\exists x \in [0,\infty)$ s.t. $f(x) > 1 - \epsilon$.
But I don't know how to kick off the proof, I'd like for some hints that can direct me.