Consider the function $g(x)=(1-c+f(x))^x$ where $x\in\mathbb{R}_{++}$ and $c\geq0$ but $f(x)$ does not have a closed-form representation. However, I know that $f(x)\geq0$, $f'(x)>0$ and $\lim_{x\to\infty}f(x)=c$. Therefore, $\lim_{x\to\infty}(1-c+f(x))=1$. I also know that $(1-c+f(x))\in[0,1]$.
I think that $\lim_{x\to\infty}g(x)$ should exist, and that the limit may be 0 or 1, but I'm not sure how to demonstrate this. If $g(x)$ was bounded and monotonic, I think I would be done, but the derivative of $g$ with respect to $x$ does not immediately suggest monotonicity and I'm not sure how to prove boundedness without a closed form of $f(x)$.
Any help would be much appreciated!