Question

How would you find the limit of this problem?:
The limit as x approaches 0 from the right of (cosx)^3/(2x)

Answer 1

have you learned l'hospitals rule?

Answer 2

You can't use l'hospital's rule for this one. You don't get 0/0.

Answer 3

ah... true.

Answer 4

you could manipulate...\[\frac{ \cos ^{3}x }{ 2x }=\frac{ \cos ^{2}x }{ x }\frac{ \cos x }{ 2 }=\frac{ 1- \sin ^{2}x }{ x }\frac{ \cos x }{ 2 }=\left( \frac{ 1 }{ x } -\sin x\frac{ \sin x }{ x }\right)\frac{ \cos x }{ 2 }\]

Answer 5

sin x / x -> 1, sin x -> 0 so 1/x - sin x (sin x/x) -> 1/x.

cos x / 2 -> 1/2

limit x ->0 1/2x

Answer 6

i dont see why would you do any of that since cos(0) is 1 so in power of 3 is still 1 , so 1 divided by 2*x=2*0=0 but since its from the right it is +infinity

Answer 7

^ Thanks, dude. That totally slipped my mind. Saved me a lot of work.