Question

compute: [\lim_{x\to 0}\frac{\sqrt{\cos(x)}-\cos(x)}{x^2}]

Answer 1

\[\lim_{x\to 0}\frac{\sqrt{\cos(x)}-\cos(x)}{x^2}\]??

Answer 2

yup, that one :) i just learned how to write it :)

Answer 3

do you know L'hopital rule?

Answer 4

just apply.

Answer 5

we're not supposed to use it

Answer 6

if it is have that form again, re apply

Answer 7

really?

Answer 8

yeah i have seen these before
"don't use l'hopital" why? who knows

Answer 9

satellite73 here, sure you get

Answer 10

i guess you can use power series, but that is even more complicated

Answer 11

@satellite73 because we don't study it yet.

Answer 12

lol, because we haven't learned it yet just like HOA said

Answer 13

maybe you are supposed to do some trick like
\[\frac{\sqrt{\cos(x)}}{x^2}-\frac{\cos(x)}{x^2}\]

Answer 14

we need to use "squeeze" rule chain rule etc.

Answer 15

strongly agree

Answer 16

tried it but i can't seem to get rid of the \[x^{2}\]

Answer 17

no my idea is bad

Answer 18

i tried the "squeeze" principle but with no luck

Answer 19

does it help to know that the answer is \(\frac{1}{4}\)?

Answer 20

lol, no... I found the answer using wolfram but I need the whole way

Answer 21

no idea?

Answer 22

I am trying now. but no result yet. do you want to give up? if so, I have no reason to continue but if you try all your best, I am with you

Answer 23

no I don't want to give up, it's math :)

Answer 24

i am not good as others, I have just one thing: stubborn!!!hihi... ok, let's try

Answer 25

thank you, I tried to multiply by \[\frac{ \sqrt{cosx}+cosx }{ \sqrt{cosx}+cosx } \]

and got to

\[\frac{ cosx - \cos^2x }{ x^2*(\sqrt{cosx}+cosx) }\]

other way was to start from:

\[-1<\cos x<\] and try to find a squeeze

Answer 26

I dont' think squeeze works, but if you factor the numerator = cosx (1-cos x) then you can break the x^2 from denominator into 2 , first term is cosx/x * (1-cosx)x * (sqr....) .

Answer 27

do you know the formula lim sinx/x =1 and lim (cos x -1)/x =0?

Answer 28

yup

Answer 29

ok, let try, you go on your way, i am on mine. if we stuck, you must wait for other's help.

Answer 30

ok, thank you

Answer 31

i've been trying to solve this allday, I think i'm missing some crucial trigonometric identity

Answer 32

see if you can make some sense of this squeeze i tried

\[-1 \le cosx \le 1\]
\[-1 \le -cosx \le 1\]
\[\sqrt{cosx}-1 \le \sqrt{cosx}-cosx \le \sqrt{cosx}+1\]
\[\frac{\sqrt{cosx}-1}{x^2} \le \frac{\sqrt{cosx}-cosx}{x^2} \le \frac{\sqrt{cosx}+1}{x^2}\]

Answer 33

sorry friend, cannot help

Answer 34

@eidinovm I accidentally wrote the result here...
http://openstudy.com/study#/updates/5154969be4b0507ceba113be

Answer 35

\[
L=\lim_{x\to0}\sqrt{\cos x}\frac{1-\sqrt{\cos x}}{x^2}\times\frac{1+\sqrt{\cos x}}{1+\sqrt{\cos x}}\\
L=\lim_{x\to0}\sqrt{\cos x}\frac{1-\cos{x}}{x^2(1+\sqrt{\cos x})}\times\frac{1+\cos x}{1+\cos x}\\
L=\lim_{x\to0}\sqrt{\cos x}\frac{1-\cos^2x}{x^2(1+\sqrt{\cos x})(1+\cos x)}\\
L=\lim_{x\to0}\sqrt{\cos x}\times\left(\frac{\sin x}{x}\right)^2\times{1\over(1+\sqrt{\cos x})(1+\cos x)}\\
L=\sqrt{\cos 0}\times(1)^2\times{1\over (1+\sqrt{\cos 0})(1+\cos 0)}\\
\boxed{\Huge L={1\over 4}}
\]

Answer 36

thank you very much!!!!!