How to derive $e^x = \lim_{n \to \infty} \left( 1 + \frac x n \right)^ n$ directly from $e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^ n$?

Question

If given the definition of $ e $:

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

Using this fact alone, can it directly derive

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

for positive and negative integer $x$ and in general, for any real number $x$?

Observing $\left(1 + \frac{x}{n}\right)^n = \left(\left(1 + \frac{1}{n/x}\right)^{n/x}\right)^x$ will handle some cases, but there is more work to be done (e.g., dealing with negative $x$, verifying that $e = \lim_{y \to \infty} \left(1 + \frac{1}{y}\right)^y$ when the limit is taken over the reals and not just the integers, etc.). — angryavian, Sep 16 '19 at 02:59
mine is to directly derive it from $e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^ n$ — nonopolarity, Sep 16 '19 at 08:41
@angryavian do we have to prove separately for integer $x$ and real $x$? — nonopolarity, Sep 16 '19 at 11:09

nonopolarity · Answer 1 · 2019-09-16T11:22:31.587

Let me try to directly derive it:

$$ \begin{align} \text{since} \ e & = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^ n \\\\ e^x & = \left(\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^ n\right)^x \\ & = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^ {nx} \\\\ \text{ let } m & = nx \ \ \text{ where } x \gt 0 \\\\ \text{ so } n & = \frac m x \\\\ e^x & = \lim_{n \to \infty} \left(1 + \frac{1}{\frac m x}\right)^ {m} \\ & = \lim_{n \to \infty} \left(1 + \frac x m\right)^ {m} \\ & = \lim_{m \to \infty} \left(1 + \frac x m\right)^ {m} \ \ \text {since $m \to \infty$ as $n \to \infty$} \\ \\ \text{re-writting the above}\\ e^x & = \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^ n \\\\ \text{It is easy to see the above holds} & \text { for $x = 0$, so}\\\\ e^x & = \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^ n \ \ \ \forall x \ge 0 \\\\ \text {Now let} \ y & = -x \ \ \ \text where \ x \ge 0\\\\ e^y &= e^{-x}\\ &= \frac 1 {e^x}\\ &= \frac {1} {\lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^ n}\\ &= \frac {1} {\lim_{n \to \infty} \left(\frac n n + \frac{x}{n}\right)^ n}\\ &= \frac {1} {\lim_{n \to \infty} \left(\frac{n + x}{n}\right)^ n}\\ &= {\lim_{n \to \infty} \left(\frac n {n + x}\right)^ n}\\ \\ \text{ let } m &= n + x \text {, so the above}\\ e^y &= {\lim_{n \to \infty} \left(\frac {m - x} {m}\right)^ {m-x}}\\ &= {\lim_{m \to \infty} \left(\frac {m - x} {m}\right)^ {m-x}} \ \ \text {since $m \to \infty$ as $n \to \infty$} \\ &= {\lim_{m \to \infty} \left(1 - \frac {x} {m}\right)^ {m-x}}\\ &= {\lim_{m \to \infty} \frac{ \left(1 - \frac {x} {m}\right)^ {m}}{ \left(1 - \frac {x} {m}\right)^ {x}}}\\ \text{ Since } {\lim_{m \to \infty} \left(1 - \frac {x} {m}\right)^ {x}} \ \ & \text {is just} \ 1 \ \text{for finite} \ x \text {, so} \ \\\ e^y &= {\lim_{m \to \infty} \left(1 - \frac {x} {m}\right)^ {m}}\\ &= {\lim_{m \to \infty} \left(1 - \frac {(-y)} {m}\right)^ {m}}\\ &= {\lim_{m \to \infty} \left(1 + \frac {y} {m}\right)^ {m}} \ \ \ \forall y = -x \le 0 \\ \\ \text {Therefore, combining the case for } \ & x \ge 0 \ \text {and $y \lt 0$, }\\\\ e^x &= {\lim_{n \to \infty} \left(1 + \frac {x} {n}\right)^ {n}} \ \ \ \forall x \in \Bbb R \\ \end{align}$$

How to derive $e^x = \lim_{n \to \infty} \left( 1 + \frac x n \right)^ n$ directly from $e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^ n$?

1 Answers1