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For fun, I tried to prove the well-known exponential property $e^{a+b} = e^a e^b$ using the limit definition of the exponential function, below.

Definition. The exponential function is defined as follows.

$$e^x := \lim_{\epsilon \rightarrow 0} \left( 1 + \epsilon x \right)^{1/\epsilon}$$

I was able to outline the majority of the proof, however I do not have sufficient justification to go from line $\eqref 1$ to step $\eqref 2$. What limit properties might be used to fill in the blanks? I'd prefer not to use the binomial theorem or calculus-based argument, if possible (though if an expansion like that seems necessary, that is OK!).

Proof.

$$e^{a+b} = \lim_{\epsilon \rightarrow 0} \left( 1 + \epsilon (a+b) \right)^{1/\epsilon}$$

$$e^{a+b} = \lim_{\epsilon \rightarrow 0} \left( 1 + \epsilon a + \epsilon b \right)^{1/\epsilon} \tag 1 \label 1$$

$$\vdots$$

$$e^{a+b} = \lim_{\epsilon \rightarrow 0} \left( 1 + \epsilon a + \epsilon b + \epsilon^2 ab\right)^{1/\epsilon} \tag 2 \label 2$$

$$e^{a+b} = \lim_{\epsilon \rightarrow 0} \left( (1 + \epsilon a)(1 + \epsilon b)\right)^{1/\epsilon}$$

$$e^{a+b} = \lim_{\epsilon \rightarrow 0} \left(1 + \epsilon a \right)^{1/\epsilon} \left(1 + \epsilon b\right)^{1/\epsilon}$$

$$e^{a+b} = e^a e^b$$

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  • $\begingroup$ See this answer math.stackexchange.com/a/3000717/72031 $\endgroup$ – Paramanand Singh Mar 2 '19 at 5:52
  • $\begingroup$ There's a proof using the Binomial Theorem on pages 66f. of J. M. Hyslop, Real Variable (Oliver & Boyd 1960) ... but [updated comment] there are nicer proofs in answers to near-duplicate questions that be found by following up Paramanand Singh's comment. $\endgroup$ – Calum Gilhooley Mar 2 '19 at 10:37
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Taking for granted that $e^x$ is well-defined in this way for all real $x$, it follows that for all real $x$: $$ e^x = \lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n. $$ For all real $a$ and $b$, and every positive integer $n$, \begin{align*} & \phantom{={}} \left\lvert\left(1+\frac{a}{n}+\frac{b}{n}+\frac{ab}{n^2}\right)^n \!\! - \left(1+\frac{a}{n}+\frac{b}{n}\right)^n\right\rvert \\ & = \frac{|ab|}{n^2}\left\lvert\left(1+\frac{a}{n}+\frac{b}{n}+\frac{ab}{n^2}\right)^{n-1} \!\! + \left(1+\frac{a}{n}+\frac{b}{n}+\frac{ab}{n^2}\right)^{n-2}\left(1+\frac{a}{n}+\frac{b}{n}\right) + \cdots \right. \\ & \phantom{={}} \left. \cdots + \left(1+\frac{a}{n}+\frac{b}{n}\right)^{n-1}\right\rvert \\ & \leqslant \frac{|ab|}{n}\left(1+\frac{|a|}{n}+\frac{|b|}{n}+\frac{|ab|}{n^2}\right)^{n-1} = \frac{|ab|}{n}\left(1+\frac{|a|}{n}\right)^{n-1} \left(1+\frac{|b|}{n}\right)^{n-1} \\ & \leqslant \frac{|ab|}{n}\left(1+\frac{|a|}{n}\right)^n \left(1+\frac{|b|}{n}\right)^n, \end{align*} and this tends to zero as $n$ tends to infinity, because: $$ \lim_{n\to\infty}\left(1+\frac{|a|}{n}\right)^n = e^{|a|} \quad\text{and}\quad \lim_{n\to\infty}\left(1+\frac{|b|}{n}\right)^n = e^{|b|}. $$ Therefore: \begin{align*} e^ae^b & = \lim_{n\to\infty}\left(1+\frac{a}{n}+\frac{b}{n}+\frac{ab}{n^2}\right)^n \\ & = \lim_{n\to\infty}\left[\left(\left(1+\frac{a}{n}+\frac{b}{n}+\frac{ab}{n^2}\right)^n \!\! - \left(1+\frac{a}{n}+\frac{b}{n}\right)^n\right) + \left(1+\frac{a}{n}+\frac{b}{n}\right)^n\right] \\ & = \lim_{n\to\infty}\left(\left(1+\frac{a}{n}+\frac{b}{n}+\frac{ab}{n^2}\right)^n \!\! - \left(1+\frac{a}{n}+\frac{b}{n}\right)^n\right) + \lim_{n\to\infty}\left(1+\frac{a}{n}+\frac{b}{n}\right)^n \\ & = 0 + e^{a+b} = e^{a+b}. \end{align*}

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