Why is $\lim_{x \to \infty}\space \left(1 + \frac{1}{x}\right)^x\space {\neq}\space 1$?

Question

I'm learning calculus, and I was given this information:

$$\lim_{x \to a} (f(x))^n = (\lim_{x \to a} f(x))^n$$ $$\lim_{x \to a}(f(x) + g(x)) = \lim_{x \to a} f(x) + \lim_{x \to a} g(x)$$ $$\lim_{x \to \infty}\frac{1}{x}=0$$ And I have to find this limit: $$\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x$$ Why can't I just use my first rule to get: $$\left(\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)\right)^x$$ then... $$\left(\lim_{x \to \infty} 1+\lim_{x \to \infty} \frac{1}{x}\right)^x$$ finally... $$(1 + 0)^x=1^x=1$$ I know the limit is actually equal to $e$ which is around $2.71828$, so why am I able to manipulate the limit to make it 1?

Answer 1

Unfortunately, the rule

$$\lim (f(x))^n=(\lim f(x))^n$$

Does not work against

$$\lim(f(x))^{g(x)}\ne(\lim f(x))^{g(x)}$$

Note that this makes no sense. For example, you would be claiming that

$$\lim_{x\to2}x^x\stackrel?=2^x$$

But clearly we actually have

$$\lim_{x\to2}x^x=2^2=4$$

Note that one results in a function, while the proper limit is a constant.

Assuming all of the following limits exist and we don't get an indeterminate form, we actually have

$$\lim(f(x))^{g(x)}=(\lim f(x))^{\lim g(x)}$$

However,

$$\lim_{x\to\infty}\left(1+\frac1x\right)^x=1^\infty$$

Which is an indeterminate form. A quick 'why?' can be seen from $1^x=1$, but $1.1^\infty=\infty$ and $0.9^\infty=0$.

Historically, this limit was one of the first definitions of $e$, one of Euler's constants,

$$\lim_{x\to\infty}\left(1+\frac1x\right)^x=e\approx2.71828182845904523\dots$$

Answer 2

If we look at $x=n\in \mathbb N$ a large natural number, we can use the binomial theorem to expand the bracket and get all positive terms $$\left(1+\frac 1n\right)^n=1+n\cdot \frac 1n + \text {other positive terms}\gt 2$$

Since we have sums larger than $2$ for at least some arbitrarily large $x$, the limit, if it exists, must be greater than $2$

---

Using the binomial theorem, and looking at the $n\cdot \frac 1n$ term, we see how the limit depends on the interaction of the $n$ inside the bracket and the $n$ which appears in the exponent. In your analysis you took the exponent outside the limit and lost this crucial interaction - look at the point at which you take two limits inside the bracket, and raise the result to the power $x$.

Answer 3

I think the confusion comes from the typical notation that we use in calculus. The limit notation $ \lim_\limits{x\rightarrow a}f(x) $ is used to denote the limit of a function at $a$ if it exists. Now notice that $x$ is just a "dummy" bound variable and changing it to $ \lim_\limits{y\rightarrow a}f(y) $ does nothing.

This notation is used purely for convinience instead of something like $\lim\limits_\infty f$, so that we can write expressions like $$\lim\limits_{x\rightarrow \infty}\left(1+\frac{1}{x}\right)^x$$ instead of the more verbose $$ \lim\limits_\infty f, \ \ \ \ \ \text{ where } f: \mathbb{R} \longrightarrow \mathbb{R}, \ \text{and for all } x \in \mathbb{R}, \ f(x) = \left(1+\frac{1}{x}\right)^x.$$

Now if you look at your first rule you will see that the $n$ in the power of $f(x)$ is free variable and is not bound to the limit notation and that's the only case when that rule works.

When you write down $$\left(\lim\limits_{x\rightarrow \infty}\left(1+\frac{1}{x}\right)\right)^x = \left(\lim\limits_{y\rightarrow \infty}\left(1+\frac{1}{y}\right)\right)^x $$ you actually mean $$ (\lim\limits_\infty g)^x, \ \ \ \ \ \text{ where } g: \mathbb{R} \longrightarrow \mathbb{R}, \ \text{and for all } y \in \mathbb{R}, \ g(y) = \left(1+\frac{1}{y}\right).$$ which is the limit of a completely different function $g$ risen to the power of some arbitrary number $x$ ($x$ is free variable here).

Answer 4

First the rule you have used is :

$\displaystyle \left\lbrace \begin{array}{l}\displaystyle \lim_{x\to \infty}g(x)=\alpha\\\displaystyle \lim_{X\to \alpha} f(X)=\ell\end{array}\right.\implies \displaystyle \lim_{x\to +\infty} f\left(g(x)\right)=\ell$

Now let $g:n \to1+ \dfrac{1}{n}$ and $f_n:x\to x^n$

You agree $f_1 \ne f_2\ne f_3$ ($f_1:x\to x, \quad f_2:x\to x^2....$), these functions are differente.

And the rule you have used, $f$ must be the same function.