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I want to prove $\lim \limits_{n\to\infty} \left(1+\frac{1}{n}\right)^{b_{n}} = e^{c}$ where $\lim \limits_{n\to\infty} \frac{b_{n}}{n}=c$.

I tried doing it with epsilons but I want to find some kind of argument that doesn't require this type of proof. So I want help with that.

Thanks in advance for anyone who helps.

user
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Yoyo666
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1 Answers1

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(Since it was answered in the comment, and I don't think I deserve the reputation gained with this one, I'll post this as a community wiki.)

$\displaystyle \lim_{n \to \infty} \left( 1+\frac 1 n \right)^{b_n} = \lim_{n \to \infty} \left[ \left(1+\frac 1 n\right)^n \right]^{\frac {b_n}{n}} = \exp\left( \lim_{n \to \infty} \dfrac{b_n}{n} \right) = \exp(c)=e^c.$

RDK
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    I think you meant to have $(1+\frac1n)^{b_n}=\exp\Big(\frac{b_n}{n}\log\big((1+\frac1n)^n\big)\Big)\xrightarrow{n\rightarrow}\exp(c\log e)=e^c$ – Mittens Jun 06 '23 at 02:13
  • @OliverDíaz The answer arguably just uses the continuity of $(x,y) \mapsto x^y$ in a neighborhood of $(e,c)$. – Brian Moehring Jun 06 '23 at 04:17
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    Indeed I would say simply by continuity $$\left(1+\frac{1}{n}\right)^{b_{n}}=\left[\left(1+\frac{1}{n}\right)^{n}\right]^\frac{b_{n}}n \to e^c$$ I don't like this $$\ldots = \lim_{n \to \infty} \left[ \left(1+\frac 1 n\right)^n \right]^{\frac {b_n}{n}} = \exp\left( \lim_{n \to \infty} \dfrac{b_n}{n} \right) = \ldots$$ in which we take the limit in two steps. As an alternative what suggested by Oliver Diaz is also fine. – user Jun 06 '23 at 08:47