# Limit as $n \rightarrow \infty$ of $(1+\frac{3}{n})^n$ [duplicate]

$$lim_{n \rightarrow \infty} (1+\frac{3}{n})^n$$

$lim_{n \rightarrow \infty}\ \frac{3}{n}=0$, and $lim_{n \rightarrow \infty}\ 1^n=1$, which is why I thought the above limit would evaluate to 1. The answer is apparently $e^3$. Why is this?

Any help will be greatly appreciated, thanks in advance.

## marked as duplicate by user99914, Nosrati, Lord Shark the Unknown, Hans Lundmark, Trevor GunnNov 5 '17 at 15:17

• The reason why the answer is not $1$ is because the base is slightly more than $1$ and any number more than $1$ raised to infinity certainly does not have to result in an answer equal to 1. You are not applying the Limit laws correctly. I am sure someone will type up a way as to how to calculate that limit. It is a common one – imranfat Nov 4 '17 at 1:21
• This is also helpful: math.stackexchange.com/questions/136784/… – imranfat Nov 4 '17 at 1:23
• Back to how you applied that limit: Under that rule, what about this one: $(3/4+3/4)^n$ where $n$ toes to infinity. According to your analogy, since $(3/4)^{big}=0$ the whole limit is $0+0=0$?. How about $1.5^{big}$? – imranfat Nov 4 '17 at 1:25
• The limit is of the form $1^{\infty}$ which is an indeterminate form – clark Nov 4 '17 at 1:39

HINT 1

$$e^x = \lim_{n \rightarrow \infty} \left( 1 + \frac{x}{n} \right)^n.$$

HINT 2

Alternatively you can write

$$\lim_{n \rightarrow \infty} \left( 1 + \frac{x}{n} \right)^n = \lim_{n \rightarrow \infty} e^{\log \left( 1 + \frac{x}{n} \right)^n} = \lim_{n \rightarrow \infty} e^{n \log \left( 1 + \frac{x}{n} \right)} = e^{{\lim_{n \rightarrow \infty}} n \log \left( 1 + \frac{x}{n} \right)} \dots$$

Then express

$$n \log \left( 1 + \frac{x}{n} \right) = \frac{\log \left( 1 + \frac{x}{n} \right) }{1/n}.$$

Then apply l'Hopitals rule.

Consider $$A=\left(1+\frac{3}{n}\right)^n\implies \log(A)=n\log\left(1+\frac{3}{n}\right)$$ Now, since $n$ is large, consider equivalents $$\log\left(1+\frac{3}{n}\right)\sim \frac{3}{n}\implies \log(A)\sim 3\implies A\sim e^3$$