4
$\begingroup$

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?

|cite|improve this question
$\endgroup$
  • $\begingroup$ What is the answer? $\endgroup$ – Michael McGovern Jul 21 '17 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 '17 at 3:07
  • $\begingroup$ Same reason $\lim_{n\to\infty} (1+1/n)^n = e \neq 1$ $\endgroup$ – Dando18 Jul 21 '17 at 3:08
  • $\begingroup$ How did you get $ (1 + 1/n)^n $ ? Can you write the solution as the answer ? $\endgroup$ – faoxis Jul 21 '17 at 3:11
  • 2
    $\begingroup$ 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 '17 at 3:37
9
$\begingroup$

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

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

|cite|improve this answer
$\endgroup$
-1
$\begingroup$

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

|cite|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.