Question

Evaluate the limit of: (1-sec^2(3x))/(6x)^2
as x goes to zero. I keep getting -1/2 and I know that is incorrect???

Answer 1

differentiate the expression twice and apply l'hopital's rule:

(-12sec(3x)^2*tan(3x)^2 - 18sec(3x)^4)/72

as x->0, this becomes:

(-12*0 - 18)/72 = -1/4

Answer 2

should be noted you're dividing the second derivative of numerator by second derivative of denominator, not just differentiating the entire expression. sorr,y should've clarified

Answer 3

Or you could rewrite it as tan^2(3x)/(6x^2). But you see when you plug in zero you get 0/0. So apply l'hospital's.
\[\frac{\frac{d^2}{dx^2}(\tan^2(3x))}{\frac{d^2}{dx^2}((6x)^2)}=\frac{18\sec^2(3x)(2\tan^2(3x)+\sec^2(3x))}{72}\].
Evaluating I get positive 1/4.

Answer 4

Refresh to fix the latex***

Answer 5

No you're right james^^. Because 1-sec^2(3x)=-tan^2(3x) I just dropped the negative.

Answer 6

Yup :P
http://www.wolframalpha.com/input/?i=limit+of+(1-sec^2(3x))%2F(6x)^2%2Cx%2C0

Answer 7

thank you< i see where I went wrong!!!

Answer 8

yep, but my differentiation was wrong...yours was correct (just out by a negative factor)...i just lucked out because the tan(3x) term vanished :P

i should take greater care with differentiating twice

Answer 9

Its all good :P

Answer 10

how would you do this without l'hopital _ I have to do it that way for class :-(