Question

If x[n]=0.99......9(n decimal places), what is the lim as n approaces infinity?

Answer 1

any guesses?

Answer 2

1?

Answer 3

Good guess :-)

Answer 4

wow, that was totally a guess. but i can see why it makes sense

Answer 5

:)

Answer 6

but can someone still expain it to me :)

Answer 7

$$0.999...=1$$look at $$\frac{1}{3}+\frac{2}{3} =1$$ $$0.333... + 0.666...= ?$$

Answer 8

0.999.......

Answer 9

$$\frac{1}{3}=0.333...$$ $$\frac{2}{3}=0.666...$$

Answer 10

if number of nines approaces infinity \[x=0.999...\]\[10x=9.999...\]\[10x-x=9.999...-0.999...=9\]\[9x=9\]\[x=1\]

Answer 11

Or
let x=0.99999...
then 10x=9.99999...
10x-x=9.99999..-0.99999...
=9
9x=9
=> x=1