Question

Help on Limits question please

Answer 1

\[\huge \lim_{x \rightarrow 0} \frac{ x^2 }{ 2-2\cos2x } \]

Answer 2

Ooo ok let's try this...

Answer 3

\[\Large \lim_{x \rightarrow 0} \frac{ x^2 }{ 2-2\cos2x }\color{royalblue}{\left(\frac{2+2\cos2x}{2+2\cos2x}\right)}\]Leading to,\[\Large\rm \lim_{x \rightarrow 0} \frac{ x^2(2+2\cos2x) }{ 4(1-\cos^22x) }\]and then,\[\Large\rm \lim_{x \rightarrow 0} \frac{ x^2(2+2\cos2x) }{ 4(\sin^22x)}\]And then ummmmmmmm maybe we can do something from there.
Ahhhhh ok maybe I made it worse.
Thinkinggggggg

Answer 4

Aha. I did something like that too! But yea....

Answer 5

So like recall this from calc 1,\[\Large\rm \lim_{x\to0}\frac{\sin x}{x}=1\]sin x behaves like x near zero.
If we square this identity thing:\[\Large\rm \left(\lim_{x\to0}\frac{\sin x}{x}\right)^2=(1)^2\]and then,\[\Large\rm \lim_{x\to0}\frac{\sin^2 x}{x^2}=1\]And thennnnnn,\[\Large\rm \lim_{x\to0}\frac{x^2}{\sin^2x}=1\]and theeeennnnnn,\[\Large\rm \lim_{x\to0}\frac{(2x)^2}{\sin^22x}=1\]And we kind of have something like that going on, yes?

Answer 6

Yepp

Answer 7

\[\Large\rm \lim_{x \rightarrow 0} \frac{ (2x)^2(1+\cos2x) }{8(\sin^22x)}\]

Answer 8

\[\Large\rm \frac{1}{8}\color{royalblue}{\lim_{x \rightarrow 0} \frac{(2x)^2}{(\sin^22x)}}\cdot \lim_{x\to0}(1+\cos2x) \]

Answer 9

Somethingggg like that, yah?

Answer 10

Ahhh ok

Answer 11

I ran through some of those steps quicky, you might wanna check my math.
I think it will work out though!\[\Large\rm \frac{1}{8}\cdot\color{royalblue}{1}\cdot \lim_{x\to0}(1+\cos2x)\]

Answer 12

Yea I checked with wolfram and it gave 1/4 which is right

Answer 13

Ooo interesting! c:
Fun problem.

Answer 14

Thanks for your help :)

Answer 15

no probs