# Prove $e^x$ limit definition from limit definition of $e$.

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 $$\left(\lim_{n\to\infty}\left(1+\frac1n\right)^n\right)^x=\lim_{n\to\infty}\left(1+\frac xn\right)^n.$$

• 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 $x$. – Paramanand Singh Jun 28 '15 at 4:31
• 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. – John Molokach Jun 28 '15 at 10:29
• 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. – Paramanand Singh Jun 28 '15 at 10:38
• 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.' – John Molokach Jun 28 '15 at 10:41
• 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. – 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 $$\lim_{n\to\infty}\left(1+\frac1n\right)^{nx}=\lim_{u\to\infty}\left(1+\frac{x}{u}\right)^u$$