OpenStudy (anonymous):

(1-cosx)/x^2

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (dumbcow):

is this a limit question

5 years ago
OpenStudy (anonymous):

Oh, sorry, as x->0

5 years ago
OpenStudy (anonymous):

:P

5 years ago
OpenStudy (dumbcow):

have you learned derivatives yet?

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

is this making any sense?

5 years ago
OpenStudy (anonymous):

Sorry, no.

5 years ago
OpenStudy (jdoe0001):

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

5 years ago
OpenStudy (jdoe0001):

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

5 years ago
OpenStudy (dumbcow):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

have a look...

5 years ago
OpenStudy (anonymous):

does this make sense?

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

no, you get 0/0

5 years ago
OpenStudy (anonymous):

?

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

it's okay... does it make sense now?

5 years ago
OpenStudy (anonymous):

Yes, thanks!

5 years ago
OpenStudy (anonymous):

you're welcome

5 years ago
OpenStudy (anonymous):

I have tons of review to do after this

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

k, thanks.

5 years ago