2

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

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

StopReadingThisUsername
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    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 04 '17 at 01:21
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    This is also helpful: https://math.stackexchange.com/questions/136784/why-lim-limits-n-to-infty-left1-frac1n-rightn-doesnt-evaluate-to?rq=1 – imranfat Nov 04 '17 at 01:23
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    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 04 '17 at 01:25
  • The limit is of the form $1^{\infty}$ which is an indeterminate form – clark Nov 04 '17 at 01:39

2 Answers2

1

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.

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

Claude Leibovici
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