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Why does the limit below equal $2$ and not $\frac{2}{e}$?

$$\lim_{n \to \infty} \frac{2}{1+\frac{1}{n}}$$

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closed as off-topic by Jonas Meyer, Claude Leibovici, අරුණ, John, Erick Wong Mar 26 at 7:55

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15  
Because it's not $\frac{2}{(1 + \frac{1}{n})^n}$? –  Qiaochu Yuan Nov 6 '11 at 20:23
1  
When you write \lim in TeX the backslash not only (1) prevents italicization but also (2) causes $n\to\infty$ to appear directly below "lim" when it's in "display" mode (as opposed to "inline") and (3) in some cases results in proper spacing between "lim" and what follows it. (I fixed it.) –  Michael Hardy Nov 6 '11 at 21:02

3 Answers 3

up vote 6 down vote accepted

$\lim_{n\rightarrow\infty}(1+1/n)^n=e$, but $\lim_{n\rightarrow\infty}(1+1/n) =1$.

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Because you are taking the limit of $\displaystyle \frac{2}{1+\frac{1}{n}}$ and not $\displaystyle \frac{2}{\left(1+\frac{1}{n}\right)^n}$. Also, because $\lim\limits_{n\to\infty}\frac{1}{n}=0$.

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Because

$$\lim_{n\to \infty}\frac{2}{(1+\frac{1}{n})}=\frac {\lim_{n\to \infty}2}{\lim_{n\to \infty}(1+\frac{1}{n})}= \frac{2}{\lim_{n\to \infty}(1+\frac{1}{n})}=\frac{2}{1}=2$$

Note that $\lim_{n\to \infty}(1+\frac{1}{n})=1$ not $e$.

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