Question

Use L'Hopital's rule to solve:

(1-sinx)/(1+cos2x) LIM x --> pi/2

Answer 1

What's the limit?

Answer 2

sorry

Answer 3

Cool. So L'Hopital's Rule only applies to indeterminite forms when you plug the limit in for x. In this case, we have a case of 0/0

Answer 4

yes

Answer 5

With L'Hopital's rule, we can still find the limit. Take the derivative of both the numerator and the denominator, and leave them in the numerator and denominator. Then, apply the limit again.

Answer 6

i did. I got -cosx/-2sin2x

Answer 7

this is 0/0?

Answer 8

Sure. Apply it again! The fun never stops.

Answer 9

i am getting sinx/-4Cos2x = -1/4

Answer 10

@iPwnBunnies : the asked question has indeterminate form .. its 0/0 form

Answer 11

Yes, the limit is -1/4. && I did address the form was 0/0. Did I miss something?

Answer 12

Oh no, you're wrong! It's 1/4, not -1/4. Check your denominator math again lol.

Answer 13

@iPwnBunnies : sorry i read it wrong..

Answer 14

You could also just use straight algebra/trig if you preferred.
Recall \[2\cos^2(x)=1+\cos(2x)\]
so we could write...
\[\frac{1-\sin(x)}{2\cos^2(x)}=\frac{1-\sin(x)}{2(1-\sin^2(x))}=\frac{1-\sin(x)}{2(1-\sin(x))(1+\sin(x))}\]
Then those 1-sin(x) 's cancel
\[\frac{1}{2(1+\sin(x))}\]
now you can plug in the value
it still get 1/4

Answer 15

Trig subs are the worst. :|

Answer 16

but how come if i use the other method I am getting a zero? - tht's what I don't get.\[\frac{ 1-sinx }{ 1+co2x }\]
\[= \frac{ (1-sinx)(1+sinx )}{(1+ \cos2 )(1+sinx)}\]
\[\frac{ 1-\sin ^{2}x }{ 1+sinxcos2x+sinx+\cos2x } = \frac{ \cos ^{2}x }{ 1+0+1+0 } =\frac{ 0 }{ 2 } =0\]

Answer 17

or what's wrong with my method

Answer 18

cos(2*pi/2)=cos(pi)=-1 not 0
so you have
1-1+1-1 on bottom not 1+0+1+0
look at the way I did it to get rid of the 0/0 case

Answer 19

thanx @myininaya