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e (Euler's number) is defined commonly in calculus and in continuous interest as:

$\lim_{n\to∞} \left(1+\frac{1}{n}\right)^n$

Why does this limit not approach 1? Since as n approaches infinity, $\frac{1}{n}$ approaches zero, hence $\left(1+\frac{1}{n}\right)$ approaches 1, and since $1^n=1$ for all cases of n, wouldn't the limit therefore be 1?

Hans Lundmark
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joshuaheckroodt
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  • Note that in your reasoning you are considering the limit $\lim_{m\to \infty} (\lim_{n\to \infty} 1+1/n)^{1/m}$, which is not the same as the first one you wrote. – Alberto Debernardi Jul 17 '17 at 12:14
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    By the same logic, you could argue that $1/n\to0$ and $0n=0$, so $n/n=(1/n)n\to0$. – Barry Cipra Jul 17 '17 at 12:15
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    The limit is applied to the entire expression at once, not to separate pieces of the expression in steps. So you cannot take $\frac{1}{n}$ to infinity first and then take the result to the power of $n$ to infinity. – Dave Jul 17 '17 at 12:17
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    Why does the limit not approach $\infty$? Since $1+\frac1n$ is greater than $1$, and raising something greater than $1$ to an ever larger power results in something that grows quickly (in fact, exponentially) towards $\infty$. As you can see, the same argument applied two different ways yields two different results, and must therefore be faulty. – Arthur Jul 17 '17 at 12:17
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    Note that $1^\infty$ is among the famous indeterminate forms, i.e., forms where $a_n\to a$ and $b_n\to b$ does not imply $a_n\circ b_n\to a\circ b$. That being said, we have $(1+\frac1n)^n\ge 1+n\cdot \frac1n=2$ for all $n$ by Bernoulli – Hagen von Eitzen Jul 17 '17 at 12:23
  • Can be useful writing the sequence as $\left(\dfrac{n+1}{n}\right)^n$ The base decreases but the exponent increases. It's amazing to see that $\left(\dfrac{11}{10}\right)^{10}$ gives already about $2.6$ – Raffaele Jul 17 '17 at 15:32

1 Answers1

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For more or less the same reason that $$ \lim_{n\to\infty} \frac{1+n}{1+n} = 0 $$ is invalid. (After all $\frac{1}{1+n} \to 0$ as $n\to\infty$, so $(1+n) \cdot \frac{1}{1+n} \to (1+n)\cdot 0 = 0$.)

As a side note, you can show that $$a_n = \Big( 1+ \frac1n \Big)^n$$ is increasing in $n$, and since $a_1 = 2$, the limit (if it exists) is surely at least $2$.

mrf
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