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Let $(x_n)$ be a sequence in R. Show that if $\lim_{n \to \infty} x_n = 0$ if follows that $\lim_{n \to \infty} (1+\frac{x_n}{n})^n = 1$

My idea looks like the following (using the binomial theorem): $$(1+\frac{x_n}{n})^n = \sum_{k=1}^{n} {{n}\choose{k}} (\frac{x_n}{n})^k = \sum_{k=1}^{n} \frac{n!}{k! \cdot (n-k)!} (\frac{x_n}{n})^k=\sum_{k=1}^{n} \frac{n \space \cdot \space ... \space \cdot \space (n-k+1)}{k!} (\frac{x_n}{n})^k$$

How do I proceed from here? Am I somehow supposed to split the sum up and then take the limit? Can someone help me out? Thanks in advance!

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  • $\begingroup$ Are you familiar with the proposition saying that if $a_n\rightarrow \pm \infty$, then $\Big( 1+\frac{1}{a_n} \Big)^{a_n}\rightarrow e$? $\endgroup$ – Keen-ameteur Jan 2 at 12:49
  • $\begingroup$ Are you sure you don't want the limit as n goes to infinity? $\endgroup$ – Zach Jan 2 at 12:51
  • $\begingroup$ No I'm not. I know that $\lim{n \to \infty} (1+\frac{x}{n})^n=e^x$ though. However I'm not sure how to apply that here. $\endgroup$ – John D. Jan 2 at 12:52
  • $\begingroup$ Oh yes as n goes to infinity. $\endgroup$ – John D. Jan 2 at 12:52
  • $\begingroup$ $0 \le \log((1+\frac{x_n}{n})^n) = n\log(1+\frac{x_n}{n}) \le n\frac{x_n}{n} = x_n$ $\endgroup$ – mathworker21 Jan 2 at 13:27
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For $n$ sufficiently large $-\epsilon <x_n <\epsilon$ so $(1-\frac {\epsilon} n)^{n} \leq (1-\frac {x_n} n)^{n} \leq (1-\frac {\epsilon} n)^{n}$. Use squeeze theorem, the fact that $(1+\frac x n)^{n} \to e^{x}$ for any real number $x$, and then observe that $e^{-\epsilon}$ and $e^{\epsilon}$ both tend to $1$ as $ \epsilon \to 0$.

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