The fact that : $$\lim_{n\to\infty} (1+1/n)^n \ne 1$$ is conterintuitive to me.
Why this doesn't work : $\lim_{n\to\infty} 1/n = 0$, then by composition : $\lim_{n\to\infty} (1+1/n)^n = 1$ ?
Is there a calculus way and intuitive way to understand why this is false ?
To determine the limit of a sequence you must look at the whole formula, and not only one term. For example, you know that $1/n$ converges to $0$ for $n\to\infty$. Does this mean that
$$\lim_{n\to\infty}\left( \frac 1n\cdot a_n\right)=0$$
for all other sequences $a_n$? I mean, you multiply $a_n$ by something which is essentially equal to zero. So why not $a_n\cdot 1/n\to 0$? Maybe because $a_n$ grows fast enought to counteract the vanishing factor $1/n$. For example, choose $a_n=n$, then you have
$$\lim_{n\to\infty}{\frac 1n\cdot n}=\lim_{n\to\infty}1=1.$$
The reason for the sequence $(1+1/n)^n$ is similar. The term in paranthesis goes to $1$. But the exponent is growing fast enough to counteract this limiting process. It is more visible when taking any logarithm:
$$\log\left(1+\frac 1n\right)^n=n\cdot \log \left(1+\frac 1n\right).$$
The right side goes to zero because the logarithm of one is zero. But the left side is growing, and this growth is fast enough to prevent the product from reaching zero. That this is indeed the case and what this will equal in the limit, that is another story. But it is not obvious (and actually wrong) that the limit is $1$.
And how to know that it could be $e$? Use the binomial formula to expand:
$$\left(1+\frac 1n\right)^n=\sum_{k=0}^n{{n\choose k} \frac1{n^k}}=\sum_{k=0}^n\left[{\frac1{k!}\cdot \color{lightgray}{\frac{n!}{(n-k)!n^k}}}\right]$$
Ignoring the gray term gives you the series representation of $e$ which is essentially its definition. So the tricky question is: why does the gray term not make a difference?
Calculus way: Let $f(x) = (1+\frac{1}{x})^x$. Now take the limit of $\log(f(x))$ as $x \to \infty$. We get
$$ \lim_{x \to \infty} x\log(1+\frac{1}{x}) = \lim_{x\to \infty} \frac{\log(1+ \frac{1}{x})}{1/x}.$$
Notice that both the numerator and the denominator go to $0$ as $x$ goes to inifinity. By L'Hopital's Rule, this limit is equal to
$$ \lim_{x \to \infty} \frac{\frac{-1/x^2}{1 + \frac{1}{x}}}{-1/x^2} = \lim_{x \to \infty} \frac{1}{1+ \frac{1}{x}} = 1.$$
Since $\lim_{x \to \infty} \log(f(x) = 1$, it follows that $\lim_{x \to \infty} f(x) = e$.
Intuitive way: Plug values in! Notice they converge to $e$:
$f(1) = 2$
$f(2) = 2.25$
$f(3) = 2.37037...$
$f(4) = 2.44140625$
$f(5) = 2.48832$
Second intuitive way: "Multiply it out." By the Binomial Theorem,
$$ (1 + \frac{1}{n})^n = 1^n + \binom{n}{1} 1^{n-1} \frac{1}{n} + \binom{n}{2} 1^{n-2} \frac{1}{n^2} + \cdots + \frac{1}{n^n}$$
In particular, just looking at the first two terms gives the approximation
$$ (1 + \frac{1}{n})^n \geq 1^n + \binom{n}{1} 1^{n-1} \frac{1}{n} = 2$$
So the limit can't be equal to $1$. In fact, it seems to be equal to some strange number that is slightly bigger than $2$...wonder what that could be. ;)
(In general, the inequality $(1+x)^n \geq 1 + nx$ is called Bernoulli's Inequality.)
The issue is that we can't compose $\displaystyle \lim_{n\to\infty} \left(1+\dfrac 1n\right)^n = 1$ from $\displaystyle \lim_{n\to\infty} \dfrac1n = 0$. Let's see what happens:
$$\lim_{n\to\infty} \dfrac1n = 0$$
$$1 + \lim_{n\to\infty} \dfrac1n = 1$$
$$\lim_{n\to\infty}1 + \lim_{n\to\infty} \dfrac1n = 1$$
$$\lim_{n\to\infty}\left( 1+\dfrac1n \right) = 1$$
$$\left[\lim_{n\to\infty}\left( 1+\dfrac1n \right)\right]^{\color{red}{n}} = 1^{{\color{red}{n}}}$$
Now if you look closely in your book at the limit composition rules, there is no rule which allows us to bring the red $\color{red}{n}$ as an exponent inside the limit.
You argument in essence is to "separate" the two uses of the $n$s and take their limits one after another. In essence you are doing this:
$\lim_n (1 + \frac 1n)^n =$
$\lim_k \{\lim_m (1+ \frac 1m)\}^k =$
$\lim_k 1^k = 1$
which is just as logical as doing:
$\lim_n (1 + \frac 1n)^n =$
$\lim_v \lim_j f(v)^j;f(v) = 1+ \frac 1v > 1$
$\lim_v \infty = \infty$
Simple answer is both of these are wrong because $n$ is not two separate variables. It is a single one.
Or another way of putting it is that as the $\frac 1n$ goes through $\frac 12, \frac 1/256, \frac 1/1149, etc.$ at the same time then $stuff^n$ will also be going through $2, 256, 1149, etc$ throwing the values off.
....
So how do we think of $\lim_n (1 + \frac 1n)^n$?
Well, I'll be honest and say looking at it I'd have no idea. But if I can show 1) it is increasing; that is $(1 + \frac 1n)^n < (1 + \frac 1{n+1})^{n+1}$ for all natural $n$ and that 2) it is bounded above; that is $(1 + \frac 1n)^n \le k$ for some $k$, then I know by the least upper bound theorem that $\lim_n(1 + \frac 1n)^n = e$ for some value I will call $e$.
As it turns out $e$ really has no particular expression in terms of other known values.
It's a good question because it seems to be that this is a violation of "the limit of a product is the product of the limits". You are viewing this as $$(1+ \frac{1}{x})\cdot (1+ \frac{1}{x}) \cdots$$ But this is not a correct application of the theorem. The theorem applies to the binary case of $f(x)\cdot g(x)$, which you can compose any finite number of times to reduce a limit of finite products to a finite product of limits (if those limits exist).
Which is to say, the limit you're asking about is not the same as your desire to interpret it as $$\lim_{m\rightarrow \infty} (\lim_{n\rightarrow \infty} (1+\frac{1}{n})^m)$$
If you review the theorems in any texts you'll have, you'll note you can "move inside an exponent" with limits only when the exponent is fixed, for instance
$$\lim_{x\rightarrow a} x^{n} = (\lim_{x\rightarrow a}x)^n$$
This is because of the criteria that are necessary to evaluate function composition. Suppose we have functions $f$ and $g$ and we wish to evaluate a limit $\lim_{x\rightarrow a}f(g(x))$. We can perform the usual evaluation and substitution provided
If these apply, then you can use your intuitive idea of limit application.
In the case of $\lim_{n\rightarrow \infty}(1 + \frac{1}{n})^n$ if you attempt to apply such substitutions you will find you cannot do anything other than restate the limit. For instance, if $g(n)=1+\frac{1}{n}$ then $f(n)=n^{\frac{1}{n-1}}$ which is no simpler than our original limit. You could try the composition with $g(n) = \frac{1}{n}$ to give $f(n) = (1+n)^{\frac{1}{n}}$ which again is no better. This limit has, to coin a phrase, an unremovable composition. So if the limit exists, it must be found in some other fashion.
I think this has been asked before but I am not able to find a link right away so here is the answer.
While solving concrete problems in mathematics we usually follow certain rules and do manipulation on mathematical symbols based on these rules. We don't perform such manipulations based on our whims and fancies. Thus for example one may fancy that world would be very nice if $\sqrt{a+b} =\sqrt{a} +\sqrt{b} $ but we know that this is not one of the rules with which we manipulate numbers and symbols representing numbers.
Sadly when dealing with limit most beginners don't realize that there are well defined rules to handle evaluation of limits and we can't just do manipulations disregarding those rules. Specifically we know that $1/n\to 0$ and there is a rule which says that under such conditions $1+1/n\to 1$ and further if $k$ is any constant number then there is another rule which says that $(1+1/n)^{k}\to 1$. But there is no specific rule which says that $(1+1/n)^{n}\to 1$.
Next lets argue via intuition. Although intuition is not powerful enough to imagine all the intricacies of analysis/calculus, it is not too weak either. Moreover most people feel that if there is an intuitive explanation for some mathematical result then it is more understandable (my personal opinion is contrary to this but let's leave that part). The limit of a sequence is based on values of the sequence and in particular we are interested in values of the sequence for large $n$. If we use calculator to calculate values of $(1+1/n)^{n}$ for some values of $n$ (say upto $n=20$) we will be convinced that the values are nowhere near $1$ and the sequence most probably does not tend to $1$. If one has more time for experiment one can try more values of $n$ and reach the same conclusion. Thus the thought that $(1+1/n)^{n}\to 1$ is neither intuitive nor based on any formal rules of limits. Instead it is based on a desire to make things simpler than they really are even at the cost of mathematical correctness (similar to the thought that $\sqrt{a+b} =\sqrt{a} +\sqrt{b} $).
If you are aware of Bernoulli's inequality then you can instantly see that $(1+1/n)^{n}\geq 2$ and hence there is no way the sequence can tend to something smaller than $2$. This way you can formally establish that the limit in question is not $1$.
The fact that the sequence $(1+1/n)^{n}$ does tend to a limit is a non trivial result which requires completeness of real numbers. It sits on an altogether different level compared to the algebraic limits of type $$\lim_{n\to\infty} \frac{n^2-1}{3n^{2}+\sqrt{2}n+1}=\frac{1}{3}$$ and it is quite natural that a beginner will have some challenges dealing with this limit. In an introductory course one may accept that the limit exists without proof and use it as one of the important but unproven formula (after all no one bothers to find proof for Heron's formula when one is calculating area of triangles).
Once you accept the fact that $$\lim_{n\to\infty} \left(1+\frac{1}{n}\right)^{n}=e$$ it may be a fun exercise to prove that $$\lim_{n\to\infty} \left(1+\frac{x}{n}\right)^{n}=e^{x}$$ for rational $x$ using algebra of limits.
To answer the question you asked, when initially taking a limit, should you get (in this case) $1^{\infty}$, that is an indeterminate form, and so without further work you can conclude nothing, much less that the limit is one.
On the other hand, if you had say, $\lim_{n \to \infty} (1)^{n}$ (which is different than your problem), you can immediately conclude the answer is $1$.
You can also "intuitively" argue that this limit should be infinity. $1+\frac 1n$ is slightly larger than $1$, let's say it is $1.00001$, if you raise it to the power of infinity, you should get infinity... $\because$ $a^\infty\to \infty$ for all $a>1$.
This interpretation is just as wrong as yours!
If you are in a Calc 1 and still not aware of Taylor series, just accept that this limit is a constant $e$.
Otherwise, one can always expand the expression using binomial theorem and use the Taylor series for $e^x$.
