Pure Mathematics - Limits

這題不難的，就是這樣算：


圖片參考：http://i187.photobucket.com/albums/x22/cshung/7008033001778.jpg


不是很難吧～！我沒有文字要再加了。

(a)
square both side and cancel (2n-1)
rearrange the terms we have
(2n-1)(2n+1) < (2n)^2 {which is true since (2n-1)(2n+1) = (2n)^2 - 1 }
All the steps are reversible, so the inequality holds.
then
put each number 1, ... , n in the inequality and multiply them all together
we have LHS < RHS

(b)
let X be the product
X is obviously greater than 0 {since it is product of positive fraction}
also less than 1/sqrt(2n+1) [from part(a)]
taking limit as n tends to infinity
1/sqrt(2n+1) tends to 0.

Therefore, by sandwich thm, the limit of the product is 0