How to derive a tight linear upper bound of $x \rightarrow \frac{1 - \exp(-Cx)}{1 - \exp(-C)}$?

Question

let $C > 0$, I'm interested in finding tight linear upper bounds for the following function : $x \rightarrow \frac{1 - \exp(-Cx)}{1 - \exp(-C)}$. The easiest way to get a linear upper bound is to use the inequality $1 - \exp(-Cx) \le Cx$ leading to the upper bound $\frac{C}{1 - \exp(-C)}x$. Can we find a tighter upper bound? Thank you.

Answer 1

Assuming $x>0$, if you want a tighter bound, think about the simplest $[1,n]$ $P_n$ Padé approximants built around $x=0$. For example the one you wrote is $$P_0=\frac{c \,e^c }{e^c-1}x$$ But $$P_1=\frac{ c\, e^c }{e^c-1 }\frac{2x }{c x+2}$$ looks better.

Try and let me know.