Let $$ e = \lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^{n} $$ where $ n \in \mathbb{R} $.
Let $$ \exp(x) = \lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n} $$ where where $ n \in \mathbb{R} $ and $ x \in \mathbb{R} $.
How can I prove that $$ e^x = \exp(x) $$ where $ e^x $ is $ e $ raised to the power of $ x $ and $ x \in \mathbb{R} $?
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In what follows we assume that $n$ is a positive integer.
For $x\in\mathbb{Q}$ the expression $e^{x} $ is defined via processes of algebra and it's existence is justified via processes of analysis. For irrational $x$ we need a specific definition of $e^{x} $ and there are multiple approaches. One approach is to define it via the limit $$\lim_{n\to\infty} \left(1+\frac{x}{n}\right)^{n}\tag{1}$$ in question. Another is to use continuous extension of $e^{x} $ for irrational $x$ (this one is more common and Rudin popularized it heavily).
For rational values of $x$ the proof is simple. The harder approach is to prove that the limit $(1)$ above exists for all real $x$ and defines a function of $x$ commonly denoted by $\exp(x) $ and establish the property that $\exp(x+y) =\exp (x) \exp(y) $. Using this one can easily show via algebra that $\exp(x) =\{\exp(1)\} ^{x} =e^{x}$ for rational values of $x$.
The easier approach is as follows (this avoids the proof of existence of limit $(1)$ for all $x$). Let $x$ be a positive integer and then we can see that $$\lim_{n\to\infty} \left(1+\frac{x}{n}\right)^{n}=\lim_{n\to\infty}\prod_{i=1}^{x}\dfrac{\left(1+\dfrac{1}{n+i-1}\right)^{n+i-1}}{\left(1+\dfrac{1}{n+i-1}\right)^{i-1}}=\prod_{i=1}^{x}e=e^{x}$$ If $x=-y$ is a negative integer then we can see that $$\lim_{n\to\infty} \left(1-\frac{y}{n}\right)^{n}=\lim_{n\to\infty}\prod_{i=1}^{y}\dfrac{\left(1+\dfrac{1}{n-i}\right)^{-i}}{\left(1+\dfrac{1}{n-i}\right)^{n-i}}=\prod_{i=1}^{y}\frac{1}{e}=e^{-y}$$ (It is best to put some values of $x, y$ like $x=3,y=2$ and write the products in expanded form to convince of the validity of algebraic manipulation above. I have avoided writing expanded form to reduce labor of typing on smartphone). So we have proved the desired result for all integers $x$ (the case $x=0$ is trivial by the way).
Lets assume that $x=p/q$ where $p\in\mathbb{Z} $ and $q\in\mathbb{Z} ^{+} $. Then we can see that $$\lim_{n\to\infty} \left(1+\frac{x}{n}\right)^{n}=\lim_{n\to\infty}\sqrt[q]{\left(1+\frac{p}{qn}\right)^{qn}}=\sqrt[q]{e^{p}}=e^{x}$$
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Here is a simple proof I could come up with. Please review this and let me know if this proof looks good.
$$ \exp(x) = \lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n = \lim_{\frac{n}{x} \to \infty} \left( 1 + \frac{1}{\frac{n}{x}} \right)^{\frac{n}{x} \cdot x} = \lim_{t \to \infty} \left( 1 + \frac{1}{t} \right)^{tx} = e^x $$
By definition, $$e = \lim_{n\rightarrow\infty}\left(1+\frac 1n\right)^n$$ Now replace $n$ by $\frac nx$, $$e = \lim_{\frac nx \rightarrow\infty}\left(1+\frac xn\right)^\frac nx$$ Now by power law for limits, we know that $$e = \Bigr[\lim_{\frac nx \rightarrow\infty}\left(1+\frac xn\right)^n\Bigr]^\frac1x$$
Also we know that when $\frac nx\rightarrow\infty$ and $x$ is an arbitary value, then $n\rightarrow\infty$. So replacing $\frac nx\rightarrow\infty$ by $n\rightarrow\infty$ and cancelling the power $\frac1x$ by raising both sides by the power $x$, we get $$e^x=\lim_{n\rightarrow\infty}\left(1+\frac xn\right)^n = \exp(x)$$ and the proof is complete.