From a literature, $$\frac{x}{x+e^{-x }}$$is said to be bounded with supremum $1$ and infimum $\frac{1}{1-e}$. It is also continuous and non-monotonic.
I'm having a hard time proving that these are the $\sup$ and $\inf$ through limits.
How can I obtain these values?
Take the derivative of this function $f(x)=\frac{x}{x+e^{-x}}$ to get: $$\frac{d}{dx}\frac{x}{x+e^{-x}}=\frac{(x+1)e^x}{(xe^x+1)^2}$$Set this equal to zero: $$\frac{(x+1)e^x}{(xe^x+1)^2}=(x+1)e^x=0$$$x$ could be $-1$ or the limit as $x$ approaches $-\infty$. Taking the limit as $x$ approaches $-\infty$, we get $0$. If we put $-1$, we get $$-\frac{1}{e-1}=\frac{1}{1-e}$$Which is less than zero. This gives us the infimum.
To find the supremum, we could put $\infty$ into $f'(x)$ and we get $0$. The limit of $f(x)$ approaching $\infty$ is $1$. So now we claim that $1$ is a supremum. To do this, we will make an informal proof and assume that there is an $x$ such that $f(x)>1$. But we get the inequality $x>x+\frac{1}{e^x}$ which isn't true.