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On Wikipedia, it says that $\lim_{n\to\infty}(1+\frac{1}{n})^n = e$ :

It [e] is approximately equal to 2.71828,[1] and is the limit of (1 + 1/n)n as n approaches infinity, ... (Source)

When I evaluate $(1+\frac{1}{n})^n$ for $n = 10^8$, I get approximately $2.718281798347$ which indeed is pretty close to $e$.

But when I try to "solve" the limit using the laws of limits, I get

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

because of the power law

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

because of the addition law

$$=\left(1+0\right)^n = 1$$

but that would mean that $e=1$, which is obviously not true.

What am I missing / doing wrong?

Thanks in advance.

Martin Sleziak
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Kevin
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    The first step is incorrect, because the power is also changing with $n$. – carmichael561 Dec 21 '16 at 22:18
  • Thanks for your answer. So the power law doesn't apply here? Also, is $\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n = \lim_{n\to\infty}\left(\lim_{n\to\infty}\left(1+\frac{1}{n}\right)\right)^n$ incorrect? – Kevin Dec 21 '16 at 22:23
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    The power law applies when the limit is being taken with respect to a variable which is distinct from the variable in the exponent. That is, if it were $\lim_{x\rightarrow a}(1+1/x)^n$ then it would work. – David Dec 21 '16 at 22:25
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    On the other hand, doing:\begin{align}\lim_{n\to\infty}\left(1+\frac1n\right)^n&=\left(\lim_{n\to\infty}\left(1+\frac1n\right)\right)^{\lim_{n\to\infty}n}\&=1^\infty\end{align}doesn't help because $1^\infty$ is indeterminate. – Akiva Weinberger Dec 22 '16 at 01:25
  • Also related: http://math.stackexchange.com/q/550485/9464 –  Dec 22 '16 at 02:51
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    Interestingly, this very issue was just recently raised at http://matheducators.stackexchange.com/questions/11801/frequent-calculus-error-replacing-interior-part-of-an-expression-with-its-limit. – mweiss Dec 22 '16 at 02:54
  • The first step is meaningless. What does the $n$ outside the limit mean? – anomaly Jul 26 '17 at 02:27

3 Answers3

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

This is wrong. The quantity on the left is independent of $n$ while the one on the right is not.

What the power law tells you is for some positive integer $m$, $$ \lim_{n\to\infty}a_n^m=(\lim_{n\to\infty}a_n)^m $$ if $a_n>0$ for all $n$ and $\lim_{n\to\infty}a_n$ exists.

[Added later:] Despite the incorrect reasoning mentioned above, it is worth to note that the limit $\lim_{n\to\infty}(1+1/n)^n$ is sometimes used as a definition of $e$. Therefore, unless you know why the limit is not $1$, it would be "unfair" to say that "e=1" is "obviously not true".

Also, you originally put the Euler-constant tag to your question, which is also incorrect. Since the constant $e$, sometimes called the "Euler number" is not the same as the Euler's constant.

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The so-called "power law" that you refer to, e.g. something like

If $n$ is an integer and $\lim\limits_{x\to a} f(x)$ exists, then $$\lim\limits_{x\to a}(f(x))^n = (\lim \limits_{x\to a} f(x))^n.$$

It refers specifically to a constant $n$ that does not change with the limit. Instead you have something like $$\lim\limits_{n\to\infty}(f(n))^n $$ where both the power and the inner function depends on $n$. In the specific example of $e$ we have that $$f(n) = 1+\frac1n.$$

Eff
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You can see that the limit is greater than or equal to two, simply by using the first two terms of the binomial expansion, and noting that the omitted terms are positive. i.e. for each $n\gt 1$ separately: $$\left(1+\frac 1n\right)^n=1+n\cdot\frac 1n+ \text {other terms}\gt 1+1$$

The source of the error in reasoning is expressed by others, but this simple check tells you that you are wrong.

Mark Bennet
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