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