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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.

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$[\frac{x}{x+1}]$ is not $0$? – S4M Sep 10 '12 at 14:34
@S4M: I forget for a while that $\frac{x}{x+1}\in (0,1)$ at infinty as André pointed. – Babak S. Sep 10 '12 at 14:46
Usually, "at infinity" means "in the limit" since there is not really an infinity in $\mathbb{R}$. What is true is that $\left\lfloor\frac{x}{x+1}\right\rfloor=0$ for all $x>0$ so $\lim\limits_{x\to+\infty}\left\lfloor\frac{x}{x+1}\right\rfloor=0$ even though $\lim\limits_{x\to+\infty}\frac{x}{x+1}=1$. – robjohn Sep 10 '12 at 15:00
@robjohn: Thanks. I found where I had mistaken in. – Babak S. Sep 10 '12 at 15:04
up vote 1 down vote accepted

For any positive $x$, we have $\left\lfloor \dfrac{x}{x+1}\right\rfloor=0$, since $0\lt \dfrac{x}{x+1}\lt 1$.

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In the current form it is pretty easy. The first summand is

$$\lim_{x\to \infty}(\frac{x}{x+1})^x=\lim_{x\to \infty}(1+\frac{1}{x})^{-x}=e^{-1}$$

by definition.

The second summand vanishes, since $\lfloor\frac{x}{x+1}\rfloor=0$ for all $x>0$.

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Thank you for the time. – Babak S. Sep 10 '12 at 14:47

you can think that$x>0$,

then $0<\frac{x}{x+1}<1$,then$[\frac{x}{x+1}]=0$.

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Thank you for the answer. :) – Babak S. Sep 10 '12 at 14:48

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