0 votes 0 votes Mathematical Logic limits calculus + – Aspirant asked Apr 12, 2017 • retagged Jul 10, 2019 by Cristine Aspirant 628 views answer comment Share Follow See all 5 Comments See all 5 5 Comments reply Show 2 previous comments Angkit commented Apr 12, 2017 reply Follow Share infinity-infinity=0 0 votes 0 votes Shubham Sharma 2 commented Apr 12, 2017 reply Follow Share Multiply with the conjugate of $\sqrt{n^2 + 2n} - [ \sqrt{n^2 + 2n} ]$ in both numerator and denominator and put n= infinite then it will result in 0. Hence A option should be correct. 1 votes 1 votes Angkit commented Apr 12, 2017 reply Follow Share ya ∞ − ∞ is indeterminate, so we rationalize the equation,and then put n, to get a 0.Sry for previous comment. 0 votes 0 votes Please log in or register to add a comment.
Best answer 2 votes 2 votes Multiply with the conjugate of $\sqrt{n^2 + 2n} - [ \sqrt{n^2 + 2n} ]$ in both numerator and denominator and put n= infinite then it will result in 0. Hence A option should be correct. Shubham Sharma 2 answered Apr 12, 2017 • selected Apr 13, 2017 by Akriti sood Shubham Sharma 2 comment Share Follow See all 2 Comments See all 2 2 Comments reply Akriti sood commented Apr 13, 2017 reply Follow Share can u show calculation? 0 votes 0 votes Shubham Sharma 2 commented Apr 13, 2017 reply Follow Share Shown 0 votes 0 votes Please log in or register to add a comment.