# 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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Najib Idrissi, Hans Lundmark, Mathmo123, Joonas Ilmavirta, Mark Fantini
If this question can be reworded to fit the rules in the help center, please edit the question.

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

• $e^{-1}/1\ne1$. 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 $\lim_{x\to \infty}(1+\frac 1x)^x=e$