I don't understand how to find the limit of the following:
(2n^2-3n-11)/(6n^2)
as n approaches infinity. I understand I should factor 1/6 out of it and that the answer is 1/3, but I don't understand the steps to get there.|||I 'll try to solve it with L' Hospital rule,
L = lim (2n虏 - 3n - 11)/(6n虏)
......n 鈫?鈭?br>
both numerator and enumerator derive with respect to n
L = lim (4n - 3)/(12n)
......n 鈫?鈭?br>
L = lim 4/12
......n 鈫?鈭?br>
L = 1/3
- - - - - - - - - - - -- - - - - - - - - - - - - - -- - - - - - - - - - - - - - -- - - 鉁?br>
Ok, I'll show you another method to solve this problem,
L = lim (2n虏 - 3n - 11)/(6n虏)
......n 鈫?鈭?br>
divided numerator and enumerator with n虏
L = lim (2 - 3/n - 11/n虏)/6
......n 鈫?鈭?br>
L = (2 - 0 - 0)/6 = 1/3|||(2n^2-3n-11)/(6n^2) = 2n^2 / 6n^2 -3n/6n^2 - 11/6n^2
= 1/3 - 1/2n - 11/6n^2
as n approaches infinity, both 1/2n and 11/6n^2 near zero, which means your limit is 1/3
Subscribe to:
Post Comments (Atom)
No comments:
Post a Comment