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

$lim_{n \rightarrow \infty}\ \frac{3}{n}=0$, 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.

  • 2
    $\begingroup$ 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 $\endgroup$
    – imranfat
    Commented Nov 4, 2017 at 1:21
  • 1
    $\begingroup$ This is also helpful: math.stackexchange.com/questions/136784/… $\endgroup$
    – imranfat
    Commented Nov 4, 2017 at 1:23
  • 1
    $\begingroup$ 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 $1.5^{big}$? $\endgroup$
    – imranfat
    Commented Nov 4, 2017 at 1:25
  • $\begingroup$ The limit is of the form $1^{\infty}$ which is an indeterminate form $\endgroup$
    – clark
    Commented Nov 4, 2017 at 1:39

2 Answers 2



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


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 $$\log\left(1+\frac{3}{n}\right)\sim \frac{3}{n}\implies \log(A)\sim 3\implies A\sim e^3$$


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