C cormglakes New Member Joined Jan 4, 2021 Messages 18 Gender Undisclosed HSC 2021 Jan 29, 2021 #1 how do you do : limx-->0 (cos2x-1)/(x^3-x^2)
Cujo10 Member Joined Dec 15, 2017 Messages 121 Gender Male HSC 2019 Uni Grad 2024 Jan 29, 2021 #2 The hospital rule, oops I mean "L'hospital rule".
Drdusk Moderator Moderator Joined Feb 24, 2017 Messages 2,022 Location a VM Gender Male HSC 2018 Uni Grad 2023 Jan 29, 2021 #3 cormglakes said: how do you do : limx-->0 (cos2x-1)/(x^3-x^2) Click to expand... cos2x = 1-2sin^2(x) However as x -> 0 sinx approaches x. So you have -2x^2/(x^3 - x^2) Cancel out the x^2 from the top and bottom. = 2
cormglakes said: how do you do : limx-->0 (cos2x-1)/(x^3-x^2) Click to expand... cos2x = 1-2sin^2(x) However as x -> 0 sinx approaches x. So you have -2x^2/(x^3 - x^2) Cancel out the x^2 from the top and bottom. = 2
Q Qeru Well-Known Member Joined Dec 30, 2020 Messages 368 Gender Male HSC 2021 Jan 29, 2021 #4 cormglakes said: how do you do : limx-->0 (cos2x-1)/(x^3-x^2) Click to expand... The first limit approaches zero since its basically the limit of and the second limit approaches 2 Edit: dumb sign mistake which is fixed. Last edited: Jan 29, 2021
cormglakes said: how do you do : limx-->0 (cos2x-1)/(x^3-x^2) Click to expand... The first limit approaches zero since its basically the limit of and the second limit approaches 2 Edit: dumb sign mistake which is fixed.