Question

What is the limit as (cosx-1)/(2x) approaches 0?

Answer 1

are you familiar with the l'hospital rule?

Answer 2

I am not, unfortunately. I have heard of it, though.

Answer 3

its one of my fav rules...it involves calculus so if you havn't learned derivatives yet then its of no use to you :|

Answer 4

first multiply (cos x + 1) in numerator and denominator and tell me what u get ?

Answer 5

he left...yeah that was my thought too but im stuck on next step

Answer 6

sorry, my connection kinda lost....
u get -sin^2 x in numerator....

Answer 7

we all lost our connection

Answer 8

you guys are over-complicating this
first off are you allowed to use l'hospital's rule?

Answer 9

actually @SinoatrialMath doesn't know that rule, so i am guiding this sum through traditional way...

Answer 10

sorry, you were gonna prove lim x->0 (1-cosx)/x algebraically ?

Answer 11

I'm sorry, I lost connection.

Answer 12

yup, not so difficult.

Answer 13

I have never seen that, please do show @hartnn

Answer 14

oh I think I see it...

Answer 15

ok, step by step.. @SinoatrialMath did u multiply and divide by (1+cos x) ?

Answer 16

(cos^2x-1)/2x(cosx+1)

Answer 17

and whats cos^2 x - 1 ??

Answer 18

sin^2x

Answer 19

actually now I don't see it anymore, I wanna see how this turns out too

Answer 20

(sin^2x)/(-sinx) ?

Answer 21

its -sin^2 x
so u have
\(\huge\frac{-sin^2 x}{2x(cos x+1)}\)

Answer 22

Ah yes, I'm following.

Answer 23

for the record, I was just gonna separate the -(1-cosx)/x and 1/2 part...
sorry to interrupt

Answer 24

now separate ONE of the 2 sin^2 x and write it like this :
\(\huge \frac{sin x}{x} \quad \frac{-sin x}{2(cos x +1)}\)
ok with this ?

Answer 25

I understand, yes.

Answer 26

now the 1st limit is standard, and its value is 1
for 2nd limit, just put x=0 and tell me what u get ?

Answer 27

-sin(0) / (2(cos(0)+1)) = 0 / (2(1+1)) = 0 / (2(2)) = 0 / 4 = 0

Answer 28

thats right so 1st limit =1
2nd limit =0
overall limit = 1*0 = 0.
ok ?

Answer 29

Wow! Thank you very much! I would have never broken it down like that, taking out one "sinx" as well as the x in the denominator. How did you think of that (taking the x out of the denominator)?

Answer 30

let me just do mine right quick\[\lim_{x\to0}\frac{\cos x-1}{2x}=-\frac12\lim_{x\to0}\frac{1-\cos x}x\]of which I think the limits\[\lim_{x\to0}\frac{1-\cos x}x=0\]is equally known, so you get\[\lim_{x\to0}\frac{\cos x-1}{2x}=-\frac12\lim_{x\to0}\frac{1-\cos x}x=-\frac12(0)=0\]

Answer 31

I'm guessing you're quite familiar w/ using that part of the squeeze theorem (lim x-> 0 of sinx/x = 1).

Answer 32

its just practice that makes u think in right direction :)

Answer 33

indeed :)

Answer 34

Oh that was clever as well, @TuringTest - that was a very nice problem.