Question

(1-cosx)/x^2

Answer 1

I'm lost on what needs to be done to convert the bottom to x vs. x^2

Answer 2

is this a limit question

Answer 3

Oh, sorry, as x->0

Answer 4

:P

Answer 5

have you learned derivatives yet?

Answer 6

multiply top and bottom by (1+cos(x)) and rework...

Answer 7

since (1+cos(x)) ->1 as x->0, it's okay (no division by 0)

Answer 8

is this making any sense?

Answer 9

Sorry, no.

Answer 10

yes @pgpilot326 \(\bf lim_{x\rightarrow 0} \cfrac{1-cos(x)}{x^2}\\
\cfrac{1-cos(x)}{x^2} \times \cfrac{1+cos(x)}{1+cos(x)} \implies \cfrac{1^2-cos^2(x)}{x^2+x^2cos(x)} \implies \cfrac{sin^2(x)}{x^2+x^2cos(x)}\\
\cfrac{sin^2(0)}{(0)^2+x^2cos(0)}\)

Answer 11

hmmm, that would still give us a 0 at the bottom :(

Answer 12

so no l'hopitals here then...
also you will need the fact that
\[\lim_{x \rightarrow 0}\frac{\sin x}{x} = 1\]

Answer 13

first, we prpbably want one of these two forms...
\[\lim_{x \rightarrow 0}\frac{ \sin x }{ x } \text{which } = 1 \text{ or } \lim_{x \rightarrow 0}\frac{ 1-\cos x }{ x } \text{which } = 0\]
since factoring doesn't give us something nice
\[\lim_{x \rightarrow 0}\frac{1- \cos x }{ x^2 }=\lim_{x \rightarrow 0}\left( \frac{1- \cos x }{ x } \right)\frac{ 1 }{ x }=0 \times \infty \]
we want to try something else

Answer 14

\[\lim_{x \rightarrow 0}\frac{1- \cos x }{ x^2 }=\lim_{x \rightarrow 0}\frac{\left(1-\cos x \right)\left( 1+\cos x \right) }{ \left( 1+\cos x \right) x^2 }=\lim_{x \rightarrow 0}\frac{\sin ^{2} x}{ \left( 1+\cos x \right) x^2 }\]
\[=\lim_{x \rightarrow 0}\frac{\sin x }{ x }\frac{\sin x }{ x }\frac{1 }{ 1+ \cos x }=\lim_{x \rightarrow 0}\frac{\sin x }{ x }\times \lim_{x \rightarrow 0}\frac{\sin x }{ x }\times \lim_{x \rightarrow 0}\frac{1 }{ 1+ \cos x }=1 \times 1 \times 1 = 1\]

Answer 15

have a look...

Answer 16

does this make sense?

Answer 17

kinda, would you just keep it 1-cosx, and plug in?

Answer 18

no, you get 0/0

Answer 19

?

Answer 20

At the end, you got 1*1*blank.

Answer 21

Would you make it 1*1*1/(1-cosx)

Answer 22

\[\lim_{x \rightarrow 0}\frac{\sin x }{ x }\times \lim_{x \rightarrow 0}\frac{\sin x }{ x }\times \lim_{x \rightarrow 0}\frac{1 }{ 1+ \cos x }= 1 \times 1 \times 1 = 1\]

Answer 23

Ya, thet's what I was saying, lol, sorry.

Answer 24

well, 1- cos x is different from 1 + cos x as x ->0

Answer 25

Oh, sorry, meant +, my comp is lagging like crazy on openstudy.

Answer 26

it's okay...

does it make sense now?

Answer 27

Yes, thanks!

Answer 28

you're welcome

Answer 29

I have tons of review to do after this

Answer 30

it's all practice and seeing many examples. then you start to get a feel for what's needed. just keep at it and ask for help if you get stuck

Answer 31

k, thanks.