Question

Challenge

Answer 1

@electrokid @Spacelimbus
please explain this

Answer 2

*
I will get back to this in a few, I am partially distracted at the moment, but I will return in approximately 10-15 minutes to check on this.

Answer 3

k

Answer 4

what are you supposed to find now?
is there something after "a"?

Answer 5

ooh
find: \[\lim_{n\to\infty}a_n\]

Answer 6

right?

Answer 7

so, we have \[
0=\lim_{n\to\infty}|a_n|=\lim_{x\to\infty}\cases{-a_n\quad a_n<0\\
a_n\qquad a_n\ge}
\]

Answer 8

ok did u look the attachment thats the question

Answer 9

yep

Answer 10

ok thx...go further

Answer 11

so,
\[
\lim_{x\to\infty}-a_n=0=\lim_{x\to\infty}a_n
\]

Answer 12

since \(a_n\) will always lie in this range,
\[-|a_n|\le a_n\le |a_n|\]

Answer 13

hence by Squeeze thorm,
\[\lim_{x\to\infty}a_n=0\]

Answer 14

in the two posts above, the two a_n should be in the "absolute value" sign

Answer 15

ohk

Answer 16

what about absolute value theorem

Answer 17

the part that says:
\[-|a_n|\le a_n\le|a_n|\]
is the absolute value theorem.

Answer 18

got it thx...

Answer 19

http://en.wikipedia.org/wiki/Absolute_value_theorem

Answer 20

it says thats the same as "Squeeze thm"

Answer 21

lemme verify with @Spacelimbus

Answer 22

yep.