Question

Find the limit for:
\[\lim_{x \rightarrow 0} \frac{6x-\sin6x}{6x-\tan6x}\]
(will award a medal to the first to show good steps which lead to the correct answer)

Answer 1

Are you allowed to use L'Hopital's rule?

Answer 2

Yep! I don't think you can do it any other way that I can see, and it's in the 0/0 indeterminate form already. There's some trick I'm not seeing when you go to do the rule though. I keep getting a loop back to yet another indeterminate form, and a explosively growing polynomial each subsequent derivation. >_<

Answer 3

okay
gimme sec

Answer 4

np :-)

Answer 5

okay, got it

Answer 6

\[\lim_{x\to 0}\frac{6x-\sin(6x)}{6x-\tan(6x)}\]=\[\lim_{x\to 0}\frac{6-6\cos(6x)}{6-\sec^2(6x)}\]=\[\lim_{x\to 0}\frac{36\sin(6x)}{-72\sec^2(6x)\tan(6x)}\]

Answer 7

=\[\lim_{x\to0}\frac{36}{-72\sec^3(6x)}\]=\[-\frac{1}{2}\]

Answer 8

after the second application of L'Hopital's rule the trick is recognizing that \[\sec^2(6x)\tan(6x)\]=\[\sec^2(6x)\frac{\sin(6x)}{\cos(6x)}\]=\[\sec^2(6x)\sec(6x)\sin(6x)\]=\[\sec^3(6x)\sin(6x)\]

Answer 9

are the steps clear?

Answer 10

Perfect! I'd give you two medals if I could, ty. :-)

Answer 11

I have another one to ask if you'd like another medal, similar instructions. Requires a trick like this again I would presume (well not the same one, granted)

Answer 12

awesome, I am trying to get a solution without using L'Hopital's, but this is more straight forward

Answer 13

*writes steps down to follow along*

Answer 14

sure

Answer 15

post it