Here, I want to find the sum: $A=\lim_{x\to +\infty}\big(\frac{x}{x+1}\big)^x+\lim_{x\to +\infty}\big[\frac{x}{x+1}\big]^x$ where $[x]$ is the floor function of $x$.
If I take $f(x)=\frac{x}{x+1}$ and $g(x)=x$ then by using: $\alpha =\lim_{x\to +\infty}\big(f(x)-1\big)g(x)$ the first limit would be $e^{\alpha}=e^{-1}$. My problem is to consider the second limit at infinity. The answer admits that $A=1/e$ so if I assume $\frac{x}{x+1}$ tends to 1, then I have $[1]^x$ which doesn’t tend to $0$. Thanks for sharing your hints.