Exponential limit $\lim_{n\to\infty}(1 + \frac{3}{n})^n What is$\lim_{n\rightarrow\infty}(1 + \frac{3}{n})^n$? I'm a little confused on this limit. The$\frac{3}{n}$part gets smaller as$n$gets bigger, so it would really just come out to$1^n$, right? • Have you heard of the number$e$? Apr 20 '13 at 18:06 • Just think of clever way to change it in the form$(1+\frac{1}{m})^{mc}$where m is some function of$n$and$c$is a constant and use that. Apr 20 '13 at 18:07 • You may also benefit from studying the answers to this question. Apr 20 '13 at 19:07 2 Answers Hint: $$\lim_{n\to \infty} \left(1 + \frac 1n\right)^n = e\tag{1}$$ Try writing $$\dfrac{3}{n} = \dfrac{1}{n/3}$$, so the exponent $$n = 3\cdot\dfrac n3$$: $$\lim_{n\to \infty} \left(1 + \frac 1{(n/3)}\right)^{3(n/3)}$$ Putting $$m = \frac n3$$ it looks very close to $$(1)$$: $$\lim_{m\to \infty} \left(1 + \frac 1{m}\right)^{3m}$$ $$\lim_{n\to \infty} \left(1 + \frac 1{(n/3)}\right)^{3(n/3)}=\lim_{m\to \infty} \left(1 + \frac 1{m}\right)^{3m}= e^3$$ • Thanks, Amzoti...for the compliment and for cluing me in on the formatting "issue" ;-) Apr 21 '13 at 0:44 First, look at why your observation fails. First, the exponent also goes to$\infty$, so you cannot really forget about it. It would be like saying $$1=\lim \frac{n}{n}=\lim n\frac 1n =\lim n\cdot 0=\lim 0=0$$ see? For example, we agree that both$n^{-2}$and$(\log n)^{-1}$get small when$n\to\infty$; however $$\lim \left(1+\frac{1}{\log n}\right)^n\to\infty$$ while $$\lim \left(1-\frac{1}{n^2}\right)^n\to 1$$ This has all a little to do with the fact that $$\lim_{n\to \infty} \left(1 + \frac 1n\right)^n = e$$ If you haven't heard of$e$before, do a little reading about it. It turns out that$e\approx 2.718281828459045\dots$. In your case, the limit turns out to be related to$e\$, since $$\mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{3}{n}} \right)^n} = \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{3}{n}} \right)^{\frac{n}{3}3}} = \cdots$$

$${\left[ {\mathop {\lim }\limits_{n \to \infty } {{\left( {1 + \frac{3}{n}} \right)}^{\frac{n}{3}}}} \right]^3} = {e^3}$$