Question

lim Cot(2x) Sin(6x) (l'hospital's rule section)
x -> 0

Answer 1

to apply L'hopitals rule you need to put it in fraction form
\[\rightarrow \frac{\cos(2x) \sin(6x)}{\sin(2x)}\]
now differentiate, use product rule for numerator
(fg)' = f'g + fg'

Answer 2

what he said

Answer 3

take the derivative of the top.. and the bottom separately

[-2sin(2x)sin(6x) + 6cos(2x)cos(6x)] / -2cos(2x)

Answer 4

= [-sin(2x)sin(6x) + 3cos(2x)cos(6x)] / -cos(2x)

Answer 5

= 0 + 3 * 1 / - 1

Answer 6

= -3

Answer 7

actually answer is positive 3

Answer 8

but i figured it all out now, ty both

Answer 9

er yeah sorry should have been -6cos(6x)... equating to 3 srry

Answer 10

no, the deriv of sin is cos... not -cos. problem was in the denom. but moot point.. i got it!!

Answer 11

i will give you medal anyway :D

Answer 12

errrrr sorry late night i meant to say 2cos(2x) :) :) read the wrong thing

Answer 13

glad this isn't a test lol

thank ye good luck :0

Answer 14

:)