Question

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

Answer 1

0/0 L'Hopital

Answer 2

does that mean DNE? how come?

Answer 3

Im going to teach myself l'hopital tonight

Answer 4

it isn't DNE though - was wrong answer?

Answer 5

If you are very early in calculus, you can get away with saying dne, but soon you will be taught how to manipulate it and get good answer.

Answer 6

DNE was the marked incorrect

Answer 7

I didn't say it was 'dne'

Answer 8

haha: :-)

Answer 9

im trying to calc it without using l'hopital

Answer 10

You can try and manipulate in a way that the limit does not go to 0/0

Answer 11

was going to change it to sin^2(2x)

Answer 12

It still gives you 0/0.
Take derivative of top, and take derivative of bottom

Answer 13

\[\frac{\tan^2(2x)}{(6x)^2}=\frac{\frac{\sin^2(2x)}{\cos^2(2x)}}{(6x)^2}=\frac{\sin^2(2x)}{(6x)^2\cos^2(2x)}\]
\[=\frac{\sin^2(2x)}{36x^2\cos^2(2x)}=\frac{1}{36}*4*\frac{\sin(2x)\sin(2x)}{(2x)(2x)(\cos^2(2x))}\]

Answer 14

1/36*4*1*1*1=4/36=1/9

Answer 15

why do the sin and cos cross out?

Answer 16

sin2x/2x->1 as x->0
sin2x/2x->1 as x->0
1/cos^(2x)->1/1=1 as x->0

Answer 17

oh, the rules!

Answer 18

thank you! Myininaya!

Answer 19

i answered that previous one with l'hospital if you want to look at it
it is an attachment

Answer 20

without*

Answer 21

oh sweet! thanks!

Answer 22

lol i keep saying with
i mean without lol

Answer 23

thanks chag and alkileez for medals lol

Answer 24

Good work. If you haven't got to L'hopital yet, you have to get creative like this.