# Is $\lim_{x\to\infty} (0.99999…)^x=1$?

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?

• Yes, that is enough. .... well, unless you need to prove $\lim_{x\to \infty} 1^x = 1$..... Which maybe for good measure you should. – fleablood Jan 25 at 0:37
• 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 $\lim_{x\to\infty} (\lim_{n\to \infty}\sum\limits_{k=1}^n 9\cdot \frac 1{10^k})^x = 1$ – fleablood Jan 25 at 0:42

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 $$0$$

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:

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

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 $$12$$.

• 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. – Joe Jan 24 at 14:47
• it seems the same two people downvoted other answers too! – Soheil Jan 28 at 20:51
• @Soheil Yes, and sadly, my answer eventually attracted $6$ downvotes, with all but one of them unexplained. – Joe Jan 28 at 20:59
• look on the bright side! you got $11$ upvotes. – Soheil Jan 28 at 21:01
• 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. – Soheil Jan 28 at 21:03

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.