Question

find the limit [(sin^2)x]/tanx as x approaches 0.

Answer 1

cos^2x=1

Answer 2

whoops didnt see the tan x

Answer 3

rewrite as sin^2(x)/[sinx/cosx] = sin(x)cos(x)
The limit of sin(x)cos(x) can be thought of as the product of the limit of sin(x) and cos(x) separately. sin(x) approaches 0 as x apparoaches 0, and cos(x) approaches 1 as x apparoches 0, so the limit of sin(x)cos(x) = 0(1) = 0

Answer 4

sin^2x/tanx = cos2x/sec^2x= cos^2x*cos^2x= 1 using L' Hospital's rule. Can you use that rule pyeh?

Answer 5

Yes you can, but it's easier to rewrite.

Answer 6

Also, derivative of sin^2(x) = 2sin(x)cos(x)

Answer 7

why did i get 1 using that rule, and you got 0?

Answer 8

see the above post; I think you took the derivative of the numerator incorrectly.