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Just a brief "simple" question. Is $\lim_{x\to\infty} (0.99999...)^x=1$? According to this question, $0.99999...=1$, so this seems to be true.

Is that enough to prove it? It seems slightly strange to write that, but I guess in the end $0.99999...$ is merely an alternative symbol to $1$, correct?

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    $\begingroup$ Yes, that is enough. .... well, unless you need to prove $\lim_{x\to \infty} 1^x = 1$..... Which maybe for good measure you should. $\endgroup$ – fleablood Jan 25 at 0:37
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    $\begingroup$ Okay..... it DOES seem weird to write $\lim_{x\to \infty}(0.9999.....)^x$ because if on is careful to write $\lim_{x\to \infty}something\ to\ do\ with\ x$ one wouldn't be lazy enough to write $0.9999.....$. We could with this as $\lim_{x\to\infty} (\lim_{n\to \infty}\sum\limits_{k=1}^n 9\cdot \frac 1{10^k})^x = 1$ $\endgroup$ – fleablood Jan 25 at 0:42
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Yes, if you mean $0.\overline{9}$ by $0.999\ldots$ then it is true, since $$\lim_{x\to \infty} 1^x = 1$$ and, as you said, $0.\overline{9}$ , is just another way to write 1.

If you just mean a lot of nine's, then it is not true, and the limit would go to $0$

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Interpreting $0.9999\ldots$ as $\lim_{n\to\infty}(1-{1\over10^n})$, which is clearly equal to $1$, what we have is a case where limits do not interchange:

$$1=\lim_{x\to\infty}1^x=\lim_{x\to\infty}\left(\lim_{n\to\infty}\left(1-{1\over10^n}\right)\right)^x\not=\lim_{n\to\infty}\left(\lim_{x\to\infty}\left(1-{1\over10^n}\right)^x\right)=\lim_{n\to\infty}0=0$$

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If $x=y$, any truth about $x$ is a truth about $y$—both $x$ and $y$ represent the same object. Sometimes the notation we use can distract us from this fact. As Kinheadpump has already noted, since $$ \lim_{x \to \infty}1^x = 1 \, , $$ it is also true that $$ \lim_{x \to \infty}(0.\overline{9})^x=1 \, . $$ Indeed, this is actually the same statement repeated over.

Keep in mind that $0.999...$ has a precise meaning. It refers to a limit of a series: $$ \lim_{N \to \infty}\sum_{n=1}^{N}\frac{9}{10^n} = 0.9 + 0.09 + 0.009 + \ldots = 1 \, . $$ If we instead consider the series $$ S=\sum_{n=1}^{N}\frac{9}{10^n} $$ for some finite value of $N$, then $$ \lim_{x \to \infty}S^x = 0 $$ Hence, the reason that $\lim_{x \to \infty}(0.\overline{9})^x=1$ is because $0.999...$ refers to the limit of a series, rather than the sum of finitely many terms. And since the limit of this series is $1$, it follows that $0.\overline{9}=1$. You are correct in saying that $0.999...$ is merely an alternative symbol for $1$, just as $5+7$ is an alternative way of writing $12$.

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    $\begingroup$ Would either of the two people who have downvoted this post explain why they have done so? If there is something wrong, then I would like to correct it. $\endgroup$ – Joe Jan 24 at 14:47
  • $\begingroup$ it seems the same two people downvoted other answers too! $\endgroup$ – Soheil Jan 28 at 20:51
  • $\begingroup$ @Soheil Yes, and sadly, my answer eventually attracted $6$ downvotes, with all but one of them unexplained. $\endgroup$ – Joe Jan 28 at 20:59
  • $\begingroup$ look on the bright side! you got $11$ upvotes. $\endgroup$ – Soheil Jan 28 at 21:01
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    $\begingroup$ Voting system isn't very important on the site. we can see some posts which have simple approach have many upvotes on the other hand some elegant answers with few upvotes. $\endgroup$ – Soheil Jan 28 at 21:03
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Whenever you have $a=1$, since $1^x:=e^{x\ln 1}=e^0=1$, you have $a^x=1^x$ (ultimately by the axiom of equality) and thus $$ \lim_{x\to\infty}a^x=\lim_{x\to\infty}1^x=\lim_{x\to\infty}1=1 $$

Depending on how you define the set of real numbers, there are several equivalent definitions to the string $0.99999\ldots$; it is equal to $1$ by any one of the definitions.

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