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I know that the limit $\lim_{x \rightarrow \infty} (1 + \frac{1}{x})^x$ equals $e$. I tried to experiment with ways to prove this because the proof in my textbook was a little too complex for me, and I proved it this way:
We'll look at the limit $\lim_{x \rightarrow \infty} \ln((1 + \frac{1}{x})^x)$. Simplifying this (From logarithm properties), we get $\lim_{x \rightarrow \infty} x \ln(1 + \frac{1}{x})$. Now, $\lim_{x \rightarrow \infty} x = \infty$ (This is a known limit in my textbook), and $\lim_{x \rightarrow \infty} \ln(1 + \frac{1}{x}) = (\text{continuity of $\ln$ and $1 + \frac{1}{x}$}) \ln(\lim_{x \rightarrow \infty} 1 + \frac{1}{x}) = \ln(1 + \frac{1}{\infty}) = \ln(1 + 0)= \ln(1) = 0$. Now, we have the indeterminate form $0 \cdot \infty$, so we can calculate $\lim_{x \rightarrow \infty} \frac{\ln(1 + \frac{1}{x})}{\frac{1}{x}}$. This is the indeterminate form $\frac{0}{0}$, so we'll use L'hopital's rule: taking the derivative of the numerator and the denominator, we have $\lim_{x \rightarrow \infty} \frac{\frac{1}{1 + \frac{1}{x}} \cdot \frac{-1}{x^2}}{\frac{-1}{x^2}} = \lim_{x \rightarrow \infty} \frac{\frac{-1}{x^2 + x}}{\frac{-1}{x^2}} = \lim_{x \rightarrow \infty} \frac{-x^2}{-x^2 - x} = \lim_{x \rightarrow \infty} \frac{-1}{-1 - \frac{1}{x}} = 1$. So $\lim_{x \rightarrow \infty} \ln((1 + \frac{1}{x})^x) = \text{($\ln$ is continuous)} \ln(\lim_{x \rightarrow \infty} (1 + \frac{1}{x})^x) = 1$, and therefore $\lim_{x \rightarrow \infty} (1 + \frac{1}{x})^x = e$. $\blacksquare$

Does this look correct? Thanks!

User1865345
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sag0li
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    What is your definition of $e$? – Kenta S Jan 16 '23 at 16:01
  • @KentaS Euler's number: $e = lim_{n \rightarrow \infty} (1 + \frac{1}{n})^n$ – sag0li Jan 16 '23 at 16:05
  • If that's already your definition, then you have nothing to prove since your statement is true by definition – StiftungWarentest Jan 16 '23 at 16:07
  • My textbook defines it as the limit of a sequence, and this is the limit of a function – sag0li Jan 16 '23 at 16:08
  • If you can show that $(1+1/x)^x$ is increasing in $x$, then you can claim this continuous limit (w.r.t. $x$) is the same as the limit over the sequence in $n$. – angryavian Jan 16 '23 at 16:10
  • Can you explain this in a little more detail? – sag0li Jan 16 '23 at 16:12
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    @sag0li in that case I see no mistake in your proof, but i guess the most proper way of proving it would be to show $(1+1/x)^x$ is monotonously growing and $e$ is the smallest upper bound. – StiftungWarentest Jan 16 '23 at 16:17
  • Ahh okay got it. Thanks! – sag0li Jan 16 '23 at 16:26
  • Since you're looking at different proofs, maybe have a look at this: https://math.stackexchange.com/a/4495797/1070376 – calc ll Jan 16 '23 at 17:54
  • To use L'Hospital you need to differentiate, i.e. to calculate the derivative, which is a limit itself. Most classical textbooks uses the following approach: they first work out some necessary standard limits (in a pre-calculus manner) and then use those to calculate the derivatives of all elementary functions. The limit $\lim_{x\to 0}\frac{\ln(1+x)}{x}=1$ is equivalent to your limit in the question. If you use L'Hospital to calculate it you kind of running ahead of the train - it becomes a circular argument. This is why, I guess, they use another method in the book you mentioned. – A.Γ. Jan 20 '23 at 18:09

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