Convergence of exponential of monotonic function?

Question

Let $f:{\mathbb R}_+\rightarrow{\mathbb R}_+$ be an increasing continuous function. We now that $$\lim_{r\rightarrow \infty} \left(1+\frac{f(x)}{r}\right)^{r} =e^{f(x)}.$$ Then $$\lim_{r\rightarrow \infty}\left\vert 1- \frac{e^{f(x)}}{\left(1+\frac{f(x)}{r}\right)^{r}}\right\vert = 0.$$ My question is (a hint would suffice), is this convergence monotone? This is, does the sequence, for each $x>0$, $G_r = \left\vert 1- \frac{e^{f(x)}}{\left(1+\frac{f(x)}{r}\right)^{r}}\right\vert \downarrow 0$ as $r\rightarrow\infty$ (monotonically)?

Answer 1

One thing: I'm wondering whether you mean uniform convergence, because if you are occupied merely with monotonic convergence, then $f(x)$ has little significance, and one can replace it with any $c>0$.

The question boils down to the behaviour of $(1+\frac{c}{r})^r$.

Recall that $(1+\frac{1}{r})^r$ monotonically increases. Now replace $r$ with $\frac{r}{c}$. Then $(1+\frac{c}{r})^{\frac{r}{c}}$ monotonically increases too. Note that $c$ is a constant.