I seek $\lim_{x \to \infty}\lim_{r \to 0}(r+1 + O(x^{-1}))^{1/r}$. A naive approach is to swap the limits and see that $\lim_{x\to \infty} (r+1+O(x^{-1}))^{1/r} = (r+1+\lim_{x\to \infty}O(x^{-1}))^{1/r}$ for contant $r$, where $\lim_{x\to \infty} O(x^{-1}) = 0$. Thus the entire limit becomes $\lim_{r \to 0} (1+r)^{1/r} = \lim_{s \to \infty} (1 + \frac{1}{s})^s = e$, so we conclude.
However, the solutions I was given use Taylor's theorem with remainder to first bound $|O(x^{r-1})| < (r+1)rx^{r-1}$ then use this to get the bound $|O(x^{-1})| < (r+1)rx^{-1} < (r+1)r$ for $x \ge 1, r\le 1$. Then they claim this justifies swapping the limits and using my argument above.
My questions are:
- How does Taylor's theorem with remainder give the bound $|O(x^{r-1})|<(r+1)rx^{r-1}$?
- Why do we need the resulting bound on $O(x^{-1})$ to interchange the order of the limits?
- What is wrong, precisely, with the naïve initial argument?