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Prove that for all $x\in \mathbb{Z}$ the sequence $(1+\frac{x}{n})^n$ is converging and it's limit is $e^x$.

I'm not even sure how to start this one. How can I even open $(1+\frac{x}{n})$ to something I can work with? I do know $\lim(1+\frac{1}{n})^n=e$, but not sure how to get from one to the other...

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Do you know the limit $$\lim_{n\to\infty}\left(1+\frac1n\right)^n=e\;?$$ –  Brian M. Scott Dec 8 '12 at 10:26
    
Yeah I do. Though I mentioned it, my bad :p –  Nescio Dec 8 '12 at 10:27
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2 Answers 2

up vote 6 down vote accepted

HINT: Write

$$\left(1+\frac{x}n\right)^n=\left(\left(1+\frac{x}n\right)^{n/x}\right)^x?$$

and substitute $u=\dfrac{x}n$.

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hmm that was simple enough. Thank you! –  Nescio Dec 8 '12 at 10:38
    
@Nescio: You’re welcome! –  Brian M. Scott Dec 8 '12 at 10:40
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You should know (sometimes by definition) that $(1+\frac1n)^n\to e$. Consider the case $x\ne0$ (with $x=0$ being trivial).Then clearly $(1+\frac1n)^{nx}\to e^x$. This already shows $(1+\frac xn)^{n}\to e^x$ at least for the subsequence of naturals $n$ where $n$ is a multiple of $x$. For the general case, investigate the difference between $(1+\frac xn)^{n}$ and $(1+\frac 1m)^{mx}$, where $m$ is a multuple of $x$ and $m\approx \frac nx$.

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