# How to find the value of this limit? $\lim\limits_{x\to\infty}\left(\frac x{x+1}\right)^x.$ [closed]

How can I evaluate this limit?

$$\lim\limits_{x\to\infty}\left(\dfrac x{x+1}\right)^x.$$

Hm I can't take this limit. I know what I have to use, but I can't. Sorry about this. Who can do it for example?

## closed as off-topic by Najib Idrissi, Hans Lundmark, Mathmo123, Joonas Ilmavirta, Mark FantiniNov 2 '14 at 13:21

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## 4 Answers

$$\lim_{x\to\infty}\Bigl(\frac x{1+x}\Bigr)^x=\lim_{x\to\infty}\Bigl(\frac{x-1}x\Bigr)^{x-1}=\lim_{x\to\infty}\Bigl(1-\frac1x\Bigr)^{x-1}=\lim_{x\to\infty}\Bigl(1-\frac1x\Bigr)^x\Big/\Bigl(1-\frac1x\Bigr)=e^{-1}/\,1=1/e$$

• $e^{-1}/1\ne1$. Check your last step again. – Akiva Weinberger Nov 12 '14 at 11:22
• @columbus8myhw ? – user2345215 Nov 12 '14 at 15:57
• Never mind. My phone rendered the math weirdly. – Akiva Weinberger Nov 12 '14 at 16:01

Let $f(x)=\left(\frac{x}{x+1}\right)^x$. Find the limit $$\lim_{x\rightarrow\infty}\ln(f(x)).$$

Suince nobody wrote the expression I was thinking of, I will:

$$\frac x{x+1}=\frac1{\frac{x+1}x}=\frac1{1+\frac1x}\implies\left(\frac x{x+1}\right)^x=\frac1{\left(1+\frac1x\right)^x}\xrightarrow[x\to\infty]{}\frac1e$$

• This is basically multiplying with $1=\frac{1/x}{1/x}$ – Alice Ryhl Nov 18 '14 at 15:40
• Well, in fact it is exactly the same. Yes. – Timbuc Nov 18 '14 at 16:54

Notice that $(\frac{x}{x+1})^x=(1-\frac{1}{x+1})^{x+1}\cdot \frac{1}{1-\frac{1}{x+1}}$ and use $\lim_{x\to \infty}(1+\frac 1x)^x=e$