Question

Evaluate \[\lim_{\theta \rightarrow 0} \frac{1-cos \theta}{sin2\theta}\](No L'Hopital's Rule!)

Answer 1

\[\sin(2\theta) = 2\sin\theta \cos\theta\]

Answer 2

Does that work?
\[\lim_{\theta \rightarrow 0} \frac{1-cos \theta}{sin2\theta}\]\[=\lim_{\theta \rightarrow 0} \frac{1-cos \theta}{2sin\theta cos\theta}\]\[=\lim_{\theta \rightarrow 0} \frac{1}{2sin\theta cos\theta}-\frac{1}{2sin\theta}\]Doesn't look nice at all..

Answer 3

Or use the much friendlier L'Hospital!

Answer 4

And no L'Hopital's Rule is allowed.

Answer 5

@RolyPoly Look at it again dude.

Answer 6

Hmm... I'd use just the L'Hospital if I could. :|

Answer 7

Hmm!\[\rm { 1 - \cos \theta \over \sin(2\theta)} = {1 - \cos \theta \over \sqrt{1 - \cos^2(2\theta)}}\]

Answer 8

May I know how does it help?

Answer 9

I'm not sure either...

Answer 10

\[\lim_{\theta \rightarrow 0} \frac{1-cos \theta}{sin2\theta}\]\[=\lim_{\theta \rightarrow 0} \frac{1-(1-2sin^2(\theta/2))}{sin2\theta}\]\[=\lim_{\theta \rightarrow 0} \frac{2sin^2(\theta/2))}{sin2\theta}\]\[=\lim_{\theta \rightarrow 0} \frac{\frac{4sin^2(\theta/2))}{\theta/2}}{\frac{2sin2\theta}{2\theta}}\]\[=\lim_{\theta \rightarrow 0} \frac{4sin(\theta/2)}{2}\]\[=2\lim_{\theta \rightarrow 0} sin(\theta/2)\]\[=2(0)\]\[=0\]

Answer 11

Is the above correct?

Answer 12

The limit breaks down to \[\frac{1}{\sin{2x}}-\frac{1}{\sin{x}}\] if that helps?

Answer 13

I'm not quite sure if that helps.
Since that would lead to 1/0 for both terms when I take the limit directly.
And even if I multiply both numerator and denominator by x, I would get (1/0)/1 for the second term when I take the limit (similar to the first term). That is still 1/0, which is illegal.

Answer 14

What about \[\frac{1-\sin({\frac{\pi}{2}-x)}}{\sin{2x}}\] as a starting point then?

Answer 15

I'd guess \[\infty\] since the top gets to zero slower than the bottom

Answer 16

Just a minute. Would you mind telling me why my approach is not valid?

Answer 17

And the answer to this question is 0.

Answer 18

is \[1-2(\sin{\frac{x}{2}})^2 = cos{x}?\]

Answer 19

By double angle formula, I suppose it is.
\[cos2x = 1-2sin^2x\]Similarly, \[cosx = 1-2sin^2(x/2)\]

Answer 20

That was the only issue I had. Otherwise everything is good.

Answer 21

Does that make sense to you now?

Answer 22

everything u did is correct.
i would suggest a easier method , without going into half angle formula for numerator,
multiply numerator and denominator by 1+cos theta
so u get 1-cos^2 theta, in numerator.

Answer 23

Hmm.. That is..
\[\lim_{x \rightarrow 0}\frac{1-cos \theta}{sin2\theta}\]\[=\lim_{x \rightarrow 0}\frac{sin^2 \theta}{2sin\theta cos\theta (1+cos\theta)}\]\[=\lim_{x \rightarrow 0}\frac{sin \theta}{2 cos\theta (1+cos\theta)}\]\[=\frac{0}{2 (1) (1+1)}\]\[=0\]Right?

Answer 24

absolutely!

Answer 25

Wow! Thanks!!! That saves me some ink :P

Answer 26

welcome ^_^