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I am trying to simplify this limit $\lim_{n \to +\infty}\left(1 + \frac{1}{n}\right)^{nx}=e^x$ into the well-known definition of $e^x$ as:

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

I would prefer a clear explanation on how to transform the first expression into the second one without involving any other exponential limit definitions or logarithmic expressions, just using limits rules and algebra, and binomal expansion if necessary.

Thank you.

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  • $\begingroup$ "I want" is really not appreciated! $\endgroup$ – amsmath Oct 10 '19 at 17:49
  • $\begingroup$ @amsmath edited $\endgroup$ – eigenslacker Oct 10 '19 at 17:50
  • $\begingroup$ What do you exactly mean by first expression and second expression? $\endgroup$ – soobster Oct 10 '19 at 17:52
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    $\begingroup$ @soobster, the first expression is $\lim_{n \to +\infty}\left(1 + \frac{1}{n}\right)^{nx}$, the second one is $\lim_{n \to +\infty}\left(1 + \frac{x}{n}\right)^{n}$. $\endgroup$ – eigenslacker Oct 10 '19 at 17:53
  • $\begingroup$ For real, possibly irrational $x$, the question becomes a little sticky. How do you define $a^x$ when $x$ is not rational? The usual way is to define $e^{x\ln a}$, and so the requirement of not "involving any other exponential limit definitions or logarithmic expressions, just using limits rules and algebra, and binomial expansion if necessary" is not really feasible for irrational $x$. $\endgroup$ – Theo Bendit Oct 10 '19 at 18:19
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If $x>0$, substitute $y=nx\implies\frac1n=\frac xy$. Then as $n\to\infty$, we have $y\to\infty$:

$$\lim_{n\to\infty}\left(1+\frac1n\right)^{nx}=\lim_{y\to\infty}\left(1+\frac xy\right)^y$$

If $x<0$, substitute $y=-nx\implies\frac1n=-\frac xy$. Then $y\to\infty$:

$$\lim_{n\to\infty}\left(1+\frac1n\right)^{nx}=\lim_{y\to-\infty}\left(1-\frac xy\right)^{-y}$$

Now swap out the symbol $y$ for $n$ in the first case and $-n$ in the second.

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    $\begingroup$ What if $x < 0$? $\endgroup$ – amsmath Oct 10 '19 at 17:57
  • $\begingroup$ If $x<0$, $\frac{x}{n}$ would still be less than $1$. @amsmath, what case is of your concerns? $\endgroup$ – soobster Oct 10 '19 at 18:02
  • $\begingroup$ @soobster Then $y\to -\infty$. But now it's corrected anyways. $\endgroup$ – amsmath Oct 10 '19 at 18:12
  • $\begingroup$ @amsmath we see $x \to \frac{\infty}{\infty}$. So that's why we choose to represent the limit by $y \to \infty$? $\endgroup$ – eigenslacker Oct 10 '19 at 20:32
  • $\begingroup$ @gatosec We treat $x$ as a constant $\endgroup$ – user170231 Oct 11 '19 at 14:51
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Hint: add pharenthesis to the expression, remember that according to power's rule, $a^{mn}=({a^m})^n$

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  • $\begingroup$ Huh? Do you care to explain a bit more? $\endgroup$ – soobster Oct 10 '19 at 17:57
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    $\begingroup$ You know that $a_n\to e$, so $a_n^x\to e^x$. $\endgroup$ – amsmath Oct 10 '19 at 17:58
  • $\begingroup$ @amsmath what do you mean by $a_n$, what is this syntax ? $\endgroup$ – eigenslacker Oct 10 '19 at 18:00
  • $\begingroup$ Of course, I mean $a_n = (1+\frac 1n)^n$. $\endgroup$ – amsmath Oct 10 '19 at 18:00
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    $\begingroup$ I thought the question was asking how to transition from $\lim_{n \rightarrow \infty} (1+\frac{1}{n})^{nx}$ to $\lim_{n \rightarrow \infty} (1+\frac{x}{n})^{n}$? Look at the comments of this post. $\endgroup$ – soobster Oct 10 '19 at 18:06

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