Question

Can anyone help me?
http://i.imgur.com/7QgAV9j.jpg
I can't use L'hosptial's rule btw. :<

Answer 1

\[\lim_{x \rightarrow 0}\frac{ 3x^2 }{ 1-\cos^2 \frac{ x }{ 2 } }=\lim_{x \rightarrow 0}\frac{ 3x^2 }{ \sin^2 \frac{ x }{ 2 } }=3\left( \lim_{x \rightarrow 0}\frac{ x }{ \sin \frac{ x }{ 2 } } \right)^2=12\left( \lim_{x \rightarrow 0}\frac{\frac{ x }{ 2 } }{ \sin \frac{ x }{ 2 } } \right)^2=12\]

Answer 2

You might not like the way I think but..
Lets first change the look of the ugly question.
PS:Apply limit x->0 everywhere :p
We know that,
\[\large \cos2x=2\cos^2x-1=>\cos^2x = \frac{1}{2}(1+\cos2x)\]
Substituting,
\[\large \frac{3x^2}{1-\frac{1}{2}(1+\cos x)}=\frac{3x^2}{\frac{1}{2}-\frac{1}{2}\cos x}=\frac{6x^2}{1-\cos x}\]
Now from here its upto you how you solve this limit..I would prefer to use series expansion of cos x if you know it..feel free to use any one you like..
\[\frac{6x^2}{1-(1-\frac{x^2}{2}+...)}=12\]
Don't know if its correct D:

Answer 3

i kept getting 12 too but this website called wolfram alpha says its 24... :/ so im really confused

Answer 4

I'd begin by replacing the denominator with sin^2((1/2)x). Then turn the numerator into a multiple of ((1/2)x)^2. Now you have something you can evaluate using the rule that the limit of sinx/x goes to 1 (which means the limit of x/sinx also goes to one).

Answer 5

@ineptatmath
http://www.wolframalpha.com/input/?i=limit+x-%3E0+3x^2%2F%281-cos^2%28x%2F2%29%29
You forgot inputting cos " ^2 " (x/2) i guess XD if you INPUT COS(X/2) u will get 24..check it out..

Answer 6

ohhhh okay ty :D

Answer 7

yw

Answer 8

thanks everyone for the suggestions and help!! :D