# Why $\lim\limits_{n\to \infty}\left(1+\frac{1}{n}\right)^n$ doesn't evaluate to 1?

I am trying to identify what the flaw is exactly when reasoning about a limit such as the definition of $\mathbf e$:

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

Now, I know there are ways of proving this limit, such as by considering the binomial expansion of $(1+\frac{1}{n})^n$ and comparing that to the Maclaurin series of $e$. So to make this clear, I am not looking for a proof of the limit definition of $e$.

I tried searching for "limit laws/rules" but none of the rules I found described the above case. Hence, I am looking for a specific rule (or perhaps a certain perspective) that will help me realize why the above does not in fact evaluate to 1.

My train of thought is as follows: at a first glance, if I wasn't already familiar with what the limit evaluates to, I probably would evaluate the expression inside the brackets first, and then apply the limit to the power. So, $$\displaystyle\lim_{n\to\infty}\frac{1}{n}=0\quad\text{and then}$$ $$\displaystyle\lim_{n\rightarrow \infty}\left(1+0\right)^n=\lim_{n\to\infty}1^n=1$$

Why is my reasoning flawed?

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this may help. –  David Mitra Apr 25 '12 at 13:27
Sometimes it helps to look at a simpler version of the flawed reasoning, with extraneous details removed. So consider $\lim_{n\to\infty}n\cdot\frac{1}{n}$. $\lim_{n\to\infty}\frac{1}{n}=0$, and so $\lim_{n\to\infty}n\cdot\frac{1}{n}=\lim_{n\to\infty}n\cdot 0=\lim_{n\to\infty} 0=0$. –  Chris Eagle Apr 25 '12 at 13:30
You are applying the "rule" $\lim f(n)^{g(n)}=\bigl(\lim f(n)\bigr)^{\lim g(n)}$ which is unjustified. –  Andrea Mori Apr 25 '12 at 13:47
There is a widespread feeling that deciding whether $\lim_{x\to a}f(x)$ exists, and computing the limit when it exists, is the same thing as "evaluating" $f$ at $a$. There is also a widespread feeling that "$\infty$" is a number, like $123456$ but bigger. Neither is the case. But the feelings combine to lead to the belief that $\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$ is an expression much like $\left(1+\frac{1}{5}\right)^5$, and subject to the same evaluation rules. –  André Nicolas Apr 25 '12 at 14:17
$n=n$. You can't let one $n$ go to infinity without the other! What you compute (correctly) is $\lim_n (\lim_m (1+1/m)^n)$. –  mt_ Apr 25 '12 at 15:22

Your "problem" already occurs with $n^{1/n}$. If you think "of $n$ first", then the limit would be infinite. If you think "of $1/n$ first", the limit would be zero. But it's neither! (as Chris mentioned above, this already happens with products, and even with sums).
There is no justification in isolating one part from the other: each new value of $n$ obviously implies a new value of $1/n$, so there's no ground in thinking that they should behave "independently".
It's good, as you suggest, to think of the base $n$ and the exponent $1/n$ as contributing opposite effects as $n \rightarrow \infty$. We have a big base to a small power; it's not clear what we get unless we know just how big and how small the numbers are! But in your example, the effect of the exponent is strong enough to fully determine the answer; a very small exponent will give you an answer close to 1, and the limit of $n^{1/n}$ is in fact 1. But $n^{1/\log(\log n)}$ goes to infinity (the effect of the base dominates); and $n^{1/\log n}$ goes to $e$ (a compromise between the two). –  Jonas Kibelbek Apr 25 '12 at 15:16