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How to find this limit:

$$ \lim_{n\to\infty} 0.99\ldots99^{10^n},$$

$$ \text{where number of 9 is } n. $$

My solution: $$ \text{if } n\to\infty \text{ then } 0.99\ldots99 \to\ 0.(9) $$ $$ \text{because } 0.(9) = 1 \implies \lim_{n\to\infty} 0.99\ldots99^{10^n}=1^{10^\infty} = 1 $$

But my solution is wrong. Why? How can I correct it?

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  • $\begingroup$ What is the answer? $\endgroup$
    – Michael McGovern
    Jul 21, 2017 at 3:04
  • $\begingroup$ I don't know. It is a test with an input form of online course, so the solution is exactly not 1... $\endgroup$
    – faoxis
    Jul 21, 2017 at 3:07
  • $\begingroup$ Same reason $\lim_{n\to\infty} (1+1/n)^n = e \neq 1$ $\endgroup$
    – Dando18
    Jul 21, 2017 at 3:08
  • $\begingroup$ How did you get $ (1 + 1/n)^n $ ? Can you write the solution as the answer ? $\endgroup$
    – faoxis
    Jul 21, 2017 at 3:11
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    $\begingroup$ You're not just letting the number of $9$s increase; you're also raising that number to a very large power. Consider $\displaystyle \left( 1 - \frac 1 n \right)^n \to \frac 1 e \text{ as } n\to\infty.$ The form $1^\infty$ is indeterminate; i.e. if $a\to 1$ and $b\to\infty$ then $a^b$ could aproach any positive number or $0$ or $+\infty$ depending on how $a$ and $b$ depend on each other. $\endgroup$
    – Michael Hardy
    Jul 21, 2017 at 3:37

2 Answers 2

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The limit you're trying to find is

$$ \lim_{n\to\infty} (1-10^{-n})^{10^n} $$

Substitute $u \mapsto 10^n$ and you get

$$ \lim_{u\to\infty} \left(1-\frac 1 u \right)^u = \frac 1 e $$

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Your answer is wrong because 1 to the power infinity is not equal to 1 but it has unspecified value just like 0/0

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