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I was doing an exercise about limits of sequences and arrived at the following limit: $$\lim_n \left(1+\frac{1}{-n} \right)^{-n}\ \ \ \ (1)$$

We are supposed to solve the limit without using L'hopital's rule. The only limit that's similar to this one that I know is:

$$\lim_n \left(1+\frac{1}{n} \right)^n = e\ \ \ \ (2)$$

But I have no clue how to get to something similar to $(2)$ starting from (1). How can this be done?

Eduardo Magalhães
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1 Answers1

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I assume you're taking $n\to\infty$ in these limits. We have $$\lim_{n\to\infty}\left(1-\frac1n\right)^{-n}=\lim_{n\to\infty}\left(\frac n{n-1}\right)^n=\lim_{n\to\infty}\left(1+\frac1{n-1}\right)^n$$ by algebra. Clearly this equals $$\lim_{n\to\infty}\left(1+\frac1n\right)^{n+1}$$ after replacing $n$ by $n+1$. I leave it to you to proceed from here by splitting this into two pieces.

Funktorality
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    Finishing what @Funktorality has started:
    $$\lim_{n\to\infty}({1+\frac{1}{n}})^{n+1}=\lim_{n\to\infty}({1+\frac{1}{n}})^{n}\lim_{n\to\infty}({1+\frac{1}{n}})^{}$$$$\lim_{n\to\infty}({1+\frac{1}{n}})^{n}=e$$$$\lim_{n\to\infty}({1+\frac{1}{n}})^{}=1$$$$\text{So the answer is just } e $$
    (aside from that, my solution was going to be identical, so I hope you don't mind it as a comment)     – Butyl Jun 17 '22 at 23:21