How do we derive the expression for $e^x$ from $e$?

Question

Trying to go in historical order here and begin with Bernoulli's formulation for $e$:

$$e = \lim_{n \to \infty} (1 + 1/n)^n$$

How do we then make the jump to

$$e^x = \lim_{n \to \infty} (1 + x/n)^n$$

I had tried doing this:

$$e^x = (\lim_{n \to \infty} (1 + 1/n)^n)^x$$

$$e^x = \lim_{n \to \infty} (1 + 1/n)^{nx}$$

Let $m = nx$ so $n = m/x$. As $n$ goes to infinity, $m$ also goes to infinity, so:

$$e^x = \lim_{m \to \infty} (1 + x/m)^{m}$$

(although we could relabel with $n=m$ I just use $m$ to use a different one)

But I was told that I'm skipping many unproven assumptions doing this. Is there an easy way to prove what I am missing or is there an easier way to arrive at the result?

Answer 1

Firstly observe that for $y\in \mathbb{R}\quad y\to +\infty$ with $y\in(n,n+1)$

$$\left(1+\frac{1}{n+1}\right)^n=\frac{ \left(1+\frac{1}{n+1}\right)^{n+1} }{\left(1+\frac{1}{n+1}\right)}\le\left(1+\frac{1}{y}\right)^y\le\left(1+\frac{1}{n}\right)^{n+1}=\left(1+\frac{1}{n}\right)^{n}\left(1+\frac{1}{n}\right)$$

thus for squeeze theorem

$$\left(1+\frac{1}{y}\right)^y\to e$$

It easy to see that the same limit holds also for $y\in \mathbb{R}\quad y\to -\infty$.

Then note that $\forall x\in\mathbb{R}\setminus\{0\}$ fixed, since $y=\frac{n}{x}\to \pm\infty\,$ by continuity and algebraic rules for limits we have that

$$\lim_{n \to \infty}\left(1+\frac{x}{n}\right)^n=\lim_{n \to \infty}\left[\left(1+\frac{x}{n}\right)^{\frac{n}x}\right]^x=\left[\lim_{y \to \pm\infty}\left(1+\frac{1}{y}\right)^{y}\right]^x\to e^x$$

For $x=0 \implies e^0 = \lim_{n \to \infty} (1 + 0/n)^n=1$.

Thus the identity holds $\forall x\in\mathbb{R}$.

Answer 2

In what follows $n$ denotes a positive integer. This is a typical / conventional use of symbol $n$.

Using a little bit of algebra you can show that $$\lim_{n\to\infty} \left(1+\frac{x}{n}\right)^{n}=e^{x}\tag{1}$$ where $x$ is rational and $e$ is defined by $$e=\lim_{n\to\infty} \left(1+\frac{1}{n}\right)^{n}\tag{2}$$ Let's first assume that $x$ is a positive integer and then $$\left(1+\frac{x}{n}\right)^{n}=\left(\frac{n+x} {n} \right) ^n=\prod_{k=1}^{x}\left(\frac{n+k}{n+k-1}\right)^n\\=\prod_{k=1}^{x}\left(1+\frac{1}{n+k-1}\right)^{n+k-1}\left(1+\frac{1}{n+k-1}\right)^{1-k}$$ The first factor in the product on right tends to $e$ and second factor tends to $1$ so that the desired limit is $\prod_{k=1}^{x}e=e^x$. If $x=-y$ is a negative integer then we have $$\left(1+\frac{x}{n}\right)^{n}=\prod_{k=1}^{y}\dfrac{1}{\left(1+\dfrac{1}{n-k}\right)^{n-k}\left(1+\dfrac{1}{n-k}\right)^{k}}$$ and thus the desired limit is $1/e^{y}=e^x$.

For $x=0$ the result is obvious and thus we have established $(1)$ when $x$ is an integer. If $x$ is a rational number say $p/q$ where $p$ is an integer and $q$ is a positive integer then $$\left(1+\frac{x}{n}\right)^{n}=\sqrt[q]{\left(1+\frac{p}{qn}\right)^{qn}}$$ which tends to $\sqrt[q]{e^{p}} =e^x$ and our proof is complete. This step involves the continuity of function $f(t) =t^{1/q}$ and should be available to you already (this can also be proved using some algebraic manipulation).

For irrational $x$ we need a definition of irrational exponents and the proof depends on the chosen definition. The result holds even when $x$ is a complex number but the proof requires somewhat different tools (but again limited to basic algebraic manipulation).

Answer 3

Given $$ n \in \mathbb N \quad x,r \in \mathbb R \quad 0<y \in \mathbb R $$

we have $$ e = \mathop {\lim }\limits_{n\, \to \,\infty } \left( {1 + {1 \over n}} \right)^{\,n} = \mathop {\lim }\limits_{x\, \to \,\infty } \left( {1 + {1 \over x}} \right)^{\,x} $$ and then $$ \eqalign{ & e^{\,y} = \mathop {\lim }\limits_{x\, \to \,\infty } \left( {1 + {1 \over x}} \right)^{\,x\,y} = \mathop {\lim }\limits_{x\, \to \,\infty } \left( {1 + {y \over {x\,y}}} \right)^{\,x\,y} = \mathop {\lim }\limits_{x\,y\, \to \,\infty } \left( {1 + {y \over {x\,y}}} \right)^{\,x\,y} = \cr & = \mathop {\lim }\limits_{r\, \to \,\infty } \left( {1 + {y \over r}} \right)^{\,r} \cr} $$

and also $$ \eqalign{ & e^{\, - y} = {1 \over {\mathop {\lim }\limits_{x\, \to \,\infty } \left( {1 + {1 \over x}} \right)^{\,\,x\,y} }} = \mathop {\lim }\limits_{x\, \to \,\infty } \left( {1 - {1 \over {1 + x}}} \right)^{\,\,x\,y} = \mathop {\lim }\limits_{x\, \to \,\infty } \left( {1 - {1 \over {1 + x}}} \right)^{\,\,\left( {1 + x\,} \right)y - y} = \cr & = {{\mathop {\lim }\limits_{1 + x\, \to \,\infty } \left( {1 - {1 \over {1 + x}}} \right)^{\,\,\left( {1 + x\,} \right)y} } \over {\mathop {\lim }\limits_{1 + x\, \to \,\infty } \left( {1 - {1 \over {1 + x}}} \right)^{\,\,y} }} = \mathop {\lim }\limits_{r\, \to \,\infty } \left( {1 - {y \over r}} \right)^{\,\,r} \cr} $$