Is there an elementary way of proving $$e^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n,$$ given $$e=\lim_{n\to\infty}\left(1+\frac1n\right)^n,$$ without using L"Hopital's rule, Binomial Theorem, derivatives, or power series?

In other words, given the above restrictions, we want to show $$\left(\lim_{n\to\infty}\left(1+\frac1n\right)^n\right)^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n.$$

  • $\begingroup$ There are numerous questions like these on MSE and almost all of them (barring a few) leave out the essential details like the definition of $e^{x}$. Please state your definition of $e^{x}$ for all $x$. $\endgroup$ – Paramanand Singh Jun 28 '15 at 4:31
  • $\begingroup$ Should be obvious. It's on the LHS of the equation. I will add for all $x\in\mathbb{R}$ if you feel it's needed. $\endgroup$ – John Molokach Jun 28 '15 at 10:29
  • $\begingroup$ Thats the fallacy which many live with. Irrational exponents can't be taken for granted (or deemed obvious) without a much deeper analysis. I suppose you want to define $e^{x}$ as limit of $e^{x_{n}}$ where $x_{n}$ is a sequence of rationals tending to $x$. This is probably the most complicated definition of $e^{x}$ and it needs substantial work to prove many properties of $e^{x}$. See paramanands.blogspot.com/2014/05/… more for details. $\endgroup$ – Paramanand Singh Jun 28 '15 at 10:38
  • $\begingroup$ I'm aware of this issue and its complexity. If I wanted to ask about this, I'd have tagged the question differently. Think 'Pre-Calculus.' $\endgroup$ – John Molokach Jun 28 '15 at 10:41
  • $\begingroup$ I wont continue this further. But if you wish to remain within "pre-calculus" then you should not ask about "proof" rather you need to ask about "informal/non-rigorous/intuitive argument" about why such identities hold. The trouble with such proofs given in pre-calculus is that they tend to make students feel as if these are perfectly valid and rigorous and this is kind of intellectual dishonesty / fraud on part of pre-calculus texbook authors. $\endgroup$ – Paramanand Singh Jun 28 '15 at 11:03

If you accept that exponentiation is continuous, then certainly $$\left(\lim_{n\to\infty}\left(1+\frac1n\right)^n\right)^x = \lim_{n\to\infty}\left(1+\frac1n\right)^{nx}$$ But if $u=nx$, then by substitution we have $$ \lim_{n\to\infty}\left(1+\frac1n\right)^{nx}=\lim_{u\to\infty}\left(1+\frac{x}{u}\right)^u $$

  • $\begingroup$ and what if I don't accept continuity? $\endgroup$ – John Molokach Jun 27 '15 at 21:00
  • $\begingroup$ Then I would have to ask you how you're defining real exponentiation without using any calculus — the only way I know of to do it is to define rational exponentiation and extend by continuity/monotonicity. $\endgroup$ – Micah Jun 27 '15 at 21:02
  • $\begingroup$ Perfect. Thank You. This seems a well enough explanation for a precalculus student. $\endgroup$ – John Molokach Jun 27 '15 at 21:04

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