Exponential limit $\lim_{n\to\infty}(1 + \frac{3}{n})^n

Question

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?

Answer 1

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 $(1)$:

$$\lim_{m\to \infty} \left(1 + \frac 1{m}\right)^{3m}$$

$$\lim_{n\to \infty} \left(1 + \frac 1{(n/3)}\right)^{3(n/3)}=\lim_{m\to \infty} \left(1 + \frac 1{m}\right)^{3m}= e^3$$

Answer 2

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

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