Show that if $\lim_{n \to \infty} x_n = 0$ if follows that $\lim_{n \to \infty} (1+\frac{x_n}{n})^n = 1$

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!

• Are you familiar with the proposition saying that if $a_n\rightarrow \pm \infty$, then $\Big( 1+\frac{1}{a_n} \Big)^{a_n}\rightarrow e$? – Keen-ameteur Jan 2 at 12:49
• Are you sure you don't want the limit as n goes to infinity? – Zach Jan 2 at 12:51
• 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. – John D. Jan 2 at 12:52
• Oh yes as n goes to infinity. – John D. Jan 2 at 12:52
• $0 \le \log((1+\frac{x_n}{n})^n) = n\log(1+\frac{x_n}{n}) \le n\frac{x_n}{n} = x_n$ – mathworker21 Jan 2 at 13:27

For $$n$$ sufficiently large $$-\epsilon so $$(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$$.