A problem with the definition of $e^x$

Question

I am very experienced in the world of calculus, but there is a certain problem I need to solve that I can't quite get my head around. The canonical definition of $e$ is $$e=\lim_{n\to\infty}{\left(1+\frac{1}{n}\right)^n}$$ Therefore, $$e^x=\lim_{n\to\infty}{\left(1+\frac{1}{n}\right)^{nx}}$$ However, I often see this written as $$e^x=\lim_{n\to\infty}{\left(1+\frac{x}{n}\right)^{n}}$$ Why do these limits have the same value?

Answer 1

Perhaps it is better to think of this the other way around. You first treat $e^x$ as purely a notation, and then set

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

which you can then prove satisfies the two properties expected of an exponential function, namely that

$$e^0 = 1$$

and

$$e^{x + y} = e^x e^y$$

. Hence, to find its "base", simply set $x = 1$. And thus you derive the expression for $e$, and use this to define it, so that the above is now a literal exponentiation.

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ADD: Now I see you want this the "first" way around, which wasn't quite clear from how the question was written (said "a problem with the definition of $\exp$", so thus why I answered it this way).

In that case, we need another definition for exponentiation of a real base to a real exponent. In fact, such a definition - and arguably a more basic and intuitive one to start with - exists: it is this.

The exponential $b^x$ with a positive base $b$ and arbitrary real number $x$, is the unique function which satisfies the following axioms.

1. $b^0 = 1$.
2. For any two real numbers $x$ and $y$, $b^{x + y} = b^x b^y$.
3. $b^x$ is continuous in $x$.

In particular, the first two properties effectively fix the values at rational numbers $x$; the third then fixes them at irrational numbers (proof would be for a different answer; but basically it boils down to that what comes out of the first two for rationals only can be thought of as having "an uncountable number of removable singularities", and then we "remove" them all.).

(From a more sophisticated perspective, the exponential maps $x \mapsto b^x$ are the homeomorphism-isomorphisms between the additive topological group of all reals and the multiplicative topological group of positive reals.)

Once we know that a function like this exists, and then define

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

we then first have a consequence

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

. We now have to move the real power into the limit. This requires a lemma:

Lemma 1. (Continuity in the base) The function $b \mapsto b^x$, with a fixed $x$, is continuous.

(Can try adding a proof but wanna keep this mostly to exposing the theory and not filling out all the technical details.). With that we can exchange the limit out from underneath the power:

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

The next step is the following rule (again, proof omitted).

Lemma 2. For any reals $x$ and $y$ and a given base $b$, $(b^x)^y = b^{xy}$.

From that, we can go to

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

Note that this works because the arbitrary real power is defined and continuous in both arguments. We can thus then take the substitution $m := nx$ to get (this follows from the limit rule for a composite of continuous functions):

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

.

Answer 2

Better yet, let $m=nx$, so $n=m/x$, and substitute in the limit. Of course, $n\to\infty\iff m\to\infty$.