# How to find limit of this function: $\lim_{n\to\infty} 0.99\ldots99^{10^n}$?

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?

• What is the answer? Jul 21, 2017 at 3:04
• I don't know. It is a test with an input form of online course, so the solution is exactly not 1... Jul 21, 2017 at 3:07
• Same reason $\lim_{n\to\infty} (1+1/n)^n = e \neq 1$ Jul 21, 2017 at 3:08
• How did you get $(1 + 1/n)^n$ ? Can you write the solution as the answer ? Jul 21, 2017 at 3:11
• 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. Jul 21, 2017 at 3:37

$$\lim_{n\to\infty} (1-10^{-n})^{10^n}$$
Substitute $u \mapsto 10^n$ and you get
$$\lim_{u\to\infty} \left(1-\frac 1 u \right)^u = \frac 1 e$$