I havn't guessed how I should proceed. I'd like some advice. Edit: with n-->inf
W wewfy New member Joined Feb 22, 2020 Messages 8 May 26, 2020 #1 I havn't guessed how I should proceed. I'd like some advice. Edit: with n-->inf Last edited: May 26, 2020
D Deleted member 4993 Guest May 26, 2020 #2 wewfy said: View attachment 19220 I havn't guessed how I should proceed. I'd like some advice. Edit: with n-->inf Click to expand... Is it: \(\displaystyle \lim_{n \to \infty}\) or \(\displaystyle \lim_{n \to 0}\) or something else?
wewfy said: View attachment 19220 I havn't guessed how I should proceed. I'd like some advice. Edit: with n-->inf Click to expand... Is it: \(\displaystyle \lim_{n \to \infty}\) or \(\displaystyle \lim_{n \to 0}\) or something else?
Steven G Elite Member Joined Dec 30, 2014 Messages 14,405 May 26, 2020 #3 In this forum we want you to solve your problem with our help. So please post back with your work so we know what help you need.
In this forum we want you to solve your problem with our help. So please post back with your work so we know what help you need.
H HallsofIvy Elite Member Joined Jan 27, 2012 Messages 7,763 May 26, 2020 #4 Since n is typically used to represent an integer I would be inclined to assume that the limit is as n goes to infinity. My first thought was to get rid of the 5th root. \(\displaystyle (a- b)(a^4+ a^3b+ a^2b^2+ ab^3+ b^4)= a^5- b^5\) Here, \(\displaystyle a= \sqrt[5]{n^5- n^4+ 1}\) and \(\displaystyle b= n\).
Since n is typically used to represent an integer I would be inclined to assume that the limit is as n goes to infinity. My first thought was to get rid of the 5th root. \(\displaystyle (a- b)(a^4+ a^3b+ a^2b^2+ ab^3+ b^4)= a^5- b^5\) Here, \(\displaystyle a= \sqrt[5]{n^5- n^4+ 1}\) and \(\displaystyle b= n\).