limits

[cormglakes]

how do you do : limx-->0 (cos2x-1)/(x^3-x^2)

 

[Cujo10]

The hospital rule, oops I mean "L'hospital rule".

 

[Drdusk]

how do you do : limx-->0 (cos2x-1)/(x^3-x^2)

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

 

[Qeru]

how do you do : limx-->0 (cos2x-1)/(x^3-x^2)

The first limit approaches zero since its basically the limit of and the second limit approaches 2

Edit: dumb sign mistake which is fixed.

 

[Randomstudent123]