Question

lim as x goes to 0:
(tan^2(4x))/((2x)^2)

Answer 1

lol, I hope you found out how to do these without L'Hopital's rule
Again, I would advise using it, it's super effective

I'll solve it if you need it

Answer 2

haha, i am substituting. Goal later on is to learn hopital!

Answer 3

lol good :)

It shouldn't be too hard, you'll do fine
glgl

Answer 4

If you don't know l'hopital you can still do this one with the rule about sinx/x.

Answer 5

got up to 4/cos^2(4x)

Answer 6

Huh?

Answer 7

Where'd your sins go?

Answer 8

they became 1's

Answer 9

figured it out, plug in the 0 cause cos of 0 =1!

Answer 10

Fair enough. Then I think the only problem is that the 4 should be on bottom.

Answer 11

answer is 4

Answer 12

But yeah, plug in for the cos

Answer 13

thanks everyone!

Answer 14

No, it's 1/4.

Answer 15

\[{tan^2(4x)\over(2x)^2} = ({sin(4x)\over x})^2({1 \over 4 cos^2(4x)})\]
\[\implies \lim_{x\rightarrow 0} {tan^2(4x)\over(2x)^2} = \lim_{x\rightarrow 0} ({sin(4x)\over x})^2({1 \over 4 cos^2(4x)}) = {1\over 4}\]

Answer 16

no it's 4

Answer 17

:-)

Answer 18

Then you typed something wrong in the original equation.

Answer 19

teacher said so

Answer 20

no, you got it written correctly

Answer 21

You have a 4 in the denominator. Where did it go?

Answer 22

one sec polpak; I am asking the teacher. I think you might be right>

Answer 23

Yes I am.

Answer 24

:-)

Answer 25

going to have to wait until the lecture is done _ can I send it to you _ i am really interested to see the result

Answer 26

sure

Answer 27

thanks a ton for your help!