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How do I derive $$\lim_{x\to\infty}(1+1/x)^x,\quad x\in\mathbb{R}\tag{1}$$ from $$\lim_{n\to\infty}(1+1/n)^n=e,\quad n\in\mathbb{N}\tag{2}$$ ?

Note:

  • Without calculus (continuity, derivative, aso.).
  • (2) is only an idea, every method using limit, Cauchy, aso is good.
StubbornAtom
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PeptideChain
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    Why no calculus? It's so easy to prove that $(1+1/x)^x$ is monotonous that way, which immediately would give you the result. – Arthur Sep 19 '16 at 18:24
  • because it has not yet been introduced in the book at the point of the exercise – PeptideChain Sep 19 '16 at 18:25
  • tnx for the downvote – PeptideChain Sep 19 '16 at 18:26
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    That weren't me, by the way. I hope whoever downvoted tells us why. Odds are it's a knee-jerk reaction to seing your question without any of your own work, but it would've been decent of them to leave a comment explaining that. Some people think new users understand things like that automatically. – Arthur Sep 19 '16 at 18:29
  • @Arthur: I never thought you could be – PeptideChain Sep 19 '16 at 18:30
  • Related http://math.stackexchange.com/questions/1933369/how-do-i-prove-this-limit – cgiovanardi Sep 20 '16 at 00:55

1 Answers1

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Hint: say $n \le x \le n+1$ ($x$ large). Then $$ \left ( 1 +{1\over x }\right )^x \le \left( 1 + {1\over n}\right)^{n+1}.$$ Do something similar to bound the term to limit below, and use the squeeze theorem.

peter a g
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