## welshfella one year ago Need help with this one:- Find limit as x --> 0 of 2csc^2 x - (1/2) csc^2 (1/2) x.

1. welshfella

both terms tend to infinity as x approaches zero so i converted them to the form f(x)/g(x) in order to apply l'hopitals but and got it to the form (0/0). Thats as far as i got.

2. ganeshie8

i think we can try to simplify the expression using trig identities

3. ganeshie8

\begin{align} 2\csc^2 x - \frac{1}{2} \csc^2 \frac{x}{2}&=\frac{1}{2}\left(\dfrac{4}{\sin^2x}-\frac{1}{\sin^2\frac{x}{2}}\right)\\~\\ &=\frac{1}{2}\left(\dfrac{4}{4\sin^2\frac{x}{2}\cos^2\frac{x}{2}}-\frac{1}{\sin^2\frac{x}{2}}\right)\\~\\ &=\frac{1}{2}\left(\dfrac{1-\cos^2\frac{x}{2}}{\sin^2\frac{x}{2}\cos^2\frac{x}{2}}\right)\\~\\ &=\frac{1}{2}\left(\dfrac{1}{\cos^2\frac{x}{2}}\right)\\~\\ \end{align} we can take the limit now

4. welshfella

yea

5. welshfella

1/2

6. thomas5267

Not really awake. Use with caution. \begin{align*} &\phantom{=}2\csc^2(x)-\frac{1}{2}\csc^2\left(\frac{1}{2}x\right)\\ &=\frac{2}{\sin^2(x)}-\frac{1}{2\sin^2\left(\frac{1}{2}x\right)}\\ &=\frac{2}{1-\cos^2(x)}-\frac{1}{1-\cos(x)}\\ &=\frac{2}{1-\cos^2(x)}-\frac{1+\cos(x)}{1-\cos^2(x)}\\ &=\frac{1-\cos(x)}{1-\cos^2(x)}\\ &=\frac{1}{1+\cos(x)} \end{align*}

7. anonymous

Another approach using the same identity: \begin{align*} 2\csc^2x-\frac{1}{2}\csc^2\frac{x}{2}&=\frac{4}{1-\cos2x}-\frac{1}{1-\cos x}\\[1ex] &=\frac{3-4\cos x+\cos2x}{(1-\cos2x)(1-\cos x)}\\[1ex] &=\frac{2(1-\cos x)^2}{(2-2\cos^2x)(1-\cos x)} \end{align*} which gives the last line as in thomas's answer.

8. ganeshie8

Ahh that identity looks much better as it avoids the half angles/fractions

9. welshfella

i didn't know that identity on the second line @ganeshie

10. welshfella

that half angle one

11. welshfella

I searched the web to find a half angle identity for sin^ x but couldnt find one

12. ganeshie8

ohh actually it is just the familiar sin(2x) formula, i skipped several steps in between.. let me add them

13. thomas5267

$\sin^2\theta = \frac{1 - \cos 2\theta}{2}$ Copied shamelessly from Wikipedia.

14. welshfella

Ah i see - you are squaring both sides Right!

15. ganeshie8

\begin{align} 2\csc^2 x - \frac{1}{2} \csc^2 \frac{x}{2}&=\frac{1}{2}\left(\dfrac{4}{\sin^2x}-\frac{1}{\sin^2\frac{x}{2}}\right)\\~\\ &=\frac{1}{2}\left(\dfrac{4}{(\sin x)^2}-\frac{1}{\sin^2\frac{x}{2}}\right)\\~\\ &=\frac{1}{2}\left(\dfrac{4}{(2\sin\frac{x}{2}\cos\frac{x}{2})^2}-\frac{1}{\sin^2\frac{x}{2}}\right)~~\color{grey}{\because \sin(2x)=2\sin x \cos x}\\~\\ &=\frac{1}{2}\left(\dfrac{4}{4\sin^2\frac{x}{2}\cos^2\frac{x}{2}}-\frac{1}{\sin^2\frac{x}{2}}\right)\\~\\ &=\frac{1}{2}\left(\dfrac{1-\cos^2\frac{x}{2}}{\sin^2\frac{x}{2}\cos^2\frac{x}{2}}\right)\\~\\ &=\frac{1}{2}\left(\dfrac{1}{\cos^2\frac{x}{2}}\right)\\~\\ \end{align} we can take the limit now

16. welshfella

yes I get it now

17. welshfella

thanks

18. ganeshie8

np:) sometimes the answer from wolfram gives some idea on how to approach these limit problems http://www.wolframalpha.com/input/?i=lim%28x%5Cto+0%29+2csc%5E2x-1%2F2csc%5E2%28x%2F2%29