real analysis - Proving that $f(x) = e^{-x}$ is not uniformly continuous on $mathbb{R}$

I have the problem of proving that $f(x) = e^{-x}$ is not uniformly continuous on $\mathbb{R}$.

Now I know the standard procedure, but I am having trouble due to $e$ which I have never dealt with in this sort of problem.

Attempt: Assume it is uniformly continuous on $\mathbb{R}$ :

Let $\epsilon = 1$, so we have $|x-y| \lt \delta \implies |e^{-x}-e^{-y}| \lt 1$

Choose $y = x - \frac{\delta}{2}$ so $|x-y| = |x - x + \frac{\delta}{2}| = \frac{\delta}{2} \lt \delta$

Now considering $|e^{-x} - e^{-y}| = |e^{-x} - e^{-x + \frac{\delta}{2}}| = |e^{-x}(1-e^{\frac{\delta}{2}})|$

Here I am stuck because it would seem that I will get that it is uniformly continuous since that statement seems to hold for all x as x increases. If x were to get to minus inf though, I would have a contradiction, I have never taken x to be negative in these proofs, is this acceptable?

Answer

You need to take $x$ to be negative; otherwise the function is in fact uniformly continuous.