Question

using L'Hopital's rule find ....

Answer 1

\[\lim_{x \rightarrow 0} \left( \frac{ 1 }{ x } - \frac{ 1 }{ \sin2x }\right)\]

Answer 2

so, where are you stuck ?

Answer 3

first obvious step is to cross-multiply

Answer 4

i think this hint is what you need ...
\(\lim_{x \rightarrow 0} \left( \frac{ 1 }{ x } - \frac{ 1 }{ \sin2x }\right) =\lim_{x \rightarrow 0} \dfrac{\sin 2x-x}{x \times \sin2x} \\ =\lim_{x \rightarrow 0} \dfrac{\sin 2x-x}{x \times 2x[\dfrac{\sin2x}{2x}]} \)
so, evaluate sin 2x/2x without L'Hopital's first....so you get x^2 in the denominator, then you can use L'Hoital's twice (or once ?) to get your limit value.

Answer 5

can't i do anything with sin2x = 2cosxsinx?

Answer 6

i don't think that would help make it easier....
what problem was with sin 2x, same problem will exist with sin x...in the denominator...

Answer 7

ok i asked that cause this was in a test question in my uni. and when we were doing the test our prof wrote sin2x - 2cosxsinx so i was thinking she was giving us a "clue" to make our work easier??...

Answer 8

i don't see how it becomes easier here -_-

Answer 9

thank you very much @hartnn

Answer 10

your welcome ^_^
glad i could help :)

Answer 11

the answer i got was -1/2. correct?

Answer 12

after 1st derivative
2cos 2x-1 / 4x
directly put x=0
......n0, i don't think -1/2 is correct...

Answer 13

http://www.wolframalpha.com/input/?i=lim+x-%3E+0+%281%2Fx-1%2Fsin+2x%29

Answer 14

i tried again on the second derivative and got -1 :)

Answer 15

2nd derivative ??
you can only use L'Hopital's when the form is 0/0 or infinity / infinity......after 1st derivative, you don't get any of these forms..

Answer 16

so does it mean if i get a form like -inf/inf i can't use l'hopital's rule? it has to be inf/inf and -inf/-inf?

Answer 17

'-' from -inf/inf can be taken out of limit, so you can use it.

Answer 18

well i used it :D. and i got -1 :)...just not sure if my answer is correct. i need someone to check for me :(

Answer 19

no, it isn't.... see the link, it has the answer that this limit does not exist.

Answer 20

ok now i'm going to change the lim x > 0 to x> inf. i think i missed that.

Answer 21

you sure x-> infinity ? highly unlikely....especially when trigonometric terms are involved.

Answer 22

well i think i forgot. i was just trying to remember the quetions from the test room. i won't bother. i'll solve other questions. thank a lot for all the help.

Answer 23

no problem... welcome ^_^