Question

What is the lim (1-cos theta)/(2sin^2 theta) as theta approahes 0?

Answer 1

Do you know the result \[\lim_{x \rightarrow 0}sinx / x =1\]

Answer 2

yeah..thts an identity

Answer 3

\[\lim_{\Theta \rightarrow 0} (1-\cos \Theta) / (2\sin ^{2}\Theta)\]

Answer 4

So using this result you get ehe answer, \[1-\cos \theta=2 \sin ^{2}(\theta/2)\]

Answer 5

wait y are they equal

Answer 6

That an identity do you know cos 2theta=cos ^2 theta - sin^2 theta?

Answer 7

oh ok..srry i forgot

Answer 8

Ok yeah so you have sin^2 (x/2) / sin^ x (x is more easier to type than theta) so multiply and divide by x^2 ,
so you have 4 sin^2 (x/2)/(x/2)^2/ (sin^2 x/x^2) now everything except 4 becomes 1. The answer is 4. Can you tell me why?

Answer 9

wen i do it in teh calculator it says the answer is 1/4....cept im not sure why

Answer 10

Oh yeaah i made a mistake you have (x/2)^2 in the denominator right so you have to divide by 4 and not mulyipli i'm sorry the answer is 1/4. Its like pretty hard to explain over computer just write it down. You'll understand

Answer 11

If you have sin (x/2) / x as x tends to 0 you have to first make x x/2 so you get 1/2 in the denominator similarly try to solve the given problem like this only there is an x^2 so you have to mutiply and divide by 4 you will get 1/4 only.

Answer 12

can u explain it the way wolframalpha did it..i understand evrything in it ecxpet wen it applied L Hopitals

Answer 13

http://www.wolframalpha.com/input/?i=limit+x+approaches+0+of+%281-cos+x%29%2F%282sin%5E2+x%29

Answer 14

im jsut askin u to explauiin it taht way b.c its already typed mathematically on teh comp

Answer 15

Do you know L hospitals rule its just another way of solving the problem do you know L hospitals rule?

Answer 16

no

Answer 17

Ok see if you have any function such as f(x)/g(x) and both f(x) and g(x) tend to 0 then in that case you can just get the limit by differentaiting numerator and denominator eg., take sin x /x as x ->0 both numerator and denominator tend to 0 so differentiate numerator and denominator you get cos x /1. Now at x=0 cos x=1 so the net limit is 1.

Answer 18

Basically \[\lim_{f(x) \rightarrow 0\lim_{g(x) \rightarrow 0}} f(x)/g(x)= f'(x)/g'(x)\]

Answer 19

Did you understand so far?

Answer 20

yes

Answer 21

ok i get it....u diffrenitate to get it so teh limit is not 0/0

Answer 22

So now in our question 1-cos x tends to 0 as x->0 since cosx=1 and denominator also tends to 0 as sin^2 x tends to 0 as sin x=0, so differentiate both numerator and denominator

Answer 23

So now apply L hospitals rule and tell me what you get....

Answer 24

1/ 2cos (x)

Answer 25

no you get 1/4cos x check again there was already a 2 in the denomintor of the question.

Answer 26

oh yeah 1/4 (lim x->0 of 1/cos x)

Answer 27

thats 1 as cos x=1

Answer 28

yeah u get 1/4 sicne as cos x goes to 0 u get 1

Answer 29

Thanls for ur help! =)

Answer 30

No problem...