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We know that $\dfrac{x-1}{x}$ converges to $1$ from below as $x \rightarrow \infty$.

But what about the convergence of $\left(\dfrac{x-1}{x}\right)^{x}$ as $x \rightarrow \infty$? Does it converge to zero or does it have a non-zero limit?

The answer is known if there is a + in the numerator. I don't know what happens if there is a - sign instead.

shuhalo
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    This is well-known: https://math.stackexchange.com/q/1965387/42969, https://math.stackexchange.com/q/2790897/42969 – found with Approach0 – Martin R Sep 05 '21 at 11:47
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    You should know $\lim_{x\to\infty}\left(1+\frac{k}{x}\right)^x=e^k$, for all $k$. – user10354138 Sep 05 '21 at 11:49
  • If $\lim_{x\to\infty}f(x)^{g(x)}$ is of the form "$1^{\infty}$", then $\lim_{x\to\infty}f(x)^{g(x)}=\lim_{x\to\infty}\exp\bigl(g(x)(f(x)-1)\bigr)$. – Joe Sep 05 '21 at 11:52

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My answer might be a circular argument since I suspect you are currently discussing the definition of Exp. However assuming you know Exp, Log, and l'Hôpital rule the answer is just a combinatio of tricks $$\lim_{x\rightarrow \infty}\left(\frac{x-1}{x}\right)^x= \lim_{x\rightarrow \infty}\exp(x\log(1-\frac{1}{x}))$$ Now we can take the limit inside exp since its continuous. We can then use l'Hôpital to compute $$\lim_{x\rightarrow \infty}x\log(1-\frac{1}{x})=\lim_{x\rightarrow \infty}\frac{\log(1-\frac{1}{x})}{\frac{1}{x}}=\lim_{x\rightarrow \infty}\frac{x}{1-x}=-1$$ and thus the answer is $$\lim_{x\rightarrow \infty}\left(\frac{x-1}{x}\right)^x=\exp(-1)$$

inflaton1
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