Question

evaluate the limit as x goes to 0:
((1-sec^2(6x))/(8x)^2

Answer 1

I got -9/16

Answer 2

sin(6x) 1 -1
------ ---------
cos(6x) 8x 8x

Answer 3

do i see that right?

Answer 4

uncertain, I think I am right though :-)

Answer 5

sin(6x) sin(6x) 1 -1
------ ---------
cos(6x) cos(6x) 8x 8x

Answer 6

the trick is to get 6x under the sin(6x) to make it go to 1

Answer 7

multiply by 6x.6x/6x.6x and the 8x parts vanish out an x ..

Answer 8

do you have a link where I can learn l'hopital on my own?

Answer 9

I just used wikipedia and that confused me :-)

Answer 10

\[\frac{sin(6x)}{6x}\frac{sin(6x)}{6x}\frac{6}{8}\frac{6}{8}\frac{1}{cos(x)}\frac{-1}{cos(x)}\] right?

Answer 11

thanks a ton!!!

Answer 12

lhopitals is just derive the top and the bottom seperately and see what you get when you put in the limit

Answer 13

gotcha

Answer 14

oops missed my negative

Answer 15

-sin(6x)/6x sin(6x)/6x 6/8cos(6x) 6/8cos(6x)

Answer 16

yes you did :-)

Answer 17

and a -1 since (1-sec^2) is equal to -tan^2

Answer 18

but, you, you;re good!

Answer 19

i see it there now lol

Answer 20

yes, that;s what I did