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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.
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.

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welshfella
 one year ago
Best ResponseYou've already chosen the best response.1both 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.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3i think we can try to simplify the expression using trig identities

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3\[\begin{align} 2\csc^2 x  \frac{1}{2} \csc^2 \frac{x}{2}&=\frac{1}{2}\left(\dfrac{4}{\sin^2x}\frac{1}{\sin^2\frac{x}{2}}\right)\\~\\ &=\frac{1}{2}\left(\dfrac{4}{4\sin^2\frac{x}{2}\cos^2\frac{x}{2}}\frac{1}{\sin^2\frac{x}{2}}\right)\\~\\ &=\frac{1}{2}\left(\dfrac{1\cos^2\frac{x}{2}}{\sin^2\frac{x}{2}\cos^2\frac{x}{2}}\right)\\~\\ &=\frac{1}{2}\left(\dfrac{1}{\cos^2\frac{x}{2}}\right)\\~\\ \end{align}\] we can take the limit now

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1Not really awake. Use with caution. \[ \begin{align*} &\phantom{=}2\csc^2(x)\frac{1}{2}\csc^2\left(\frac{1}{2}x\right)\\ &=\frac{2}{\sin^2(x)}\frac{1}{2\sin^2\left(\frac{1}{2}x\right)}\\ &=\frac{2}{1\cos^2(x)}\frac{1}{1\cos(x)}\\ &=\frac{2}{1\cos^2(x)}\frac{1+\cos(x)}{1\cos^2(x)}\\ &=\frac{1\cos(x)}{1\cos^2(x)}\\ &=\frac{1}{1+\cos(x)} \end{align*} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Another approach using the same identity: \[\begin{align*} 2\csc^2x\frac{1}{2}\csc^2\frac{x}{2}&=\frac{4}{1\cos2x}\frac{1}{1\cos x}\\[1ex] &=\frac{34\cos x+\cos2x}{(1\cos2x)(1\cos x)}\\[1ex] &=\frac{2(1\cos x)^2}{(22\cos^2x)(1\cos x)} \end{align*}\] which gives the last line as in thomas's answer.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Ahh that identity looks much better as it avoids the half angles/fractions

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1i didn't know that identity on the second line @ganeshie

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1that half angle one

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1I searched the web to find a half angle identity for sin^ x but couldnt find one

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3ohh actually it is just the familiar sin(2x) formula, i skipped several steps in between.. let me add them

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1\[ \sin^2\theta = \frac{1  \cos 2\theta}{2} \] Copied shamelessly from Wikipedia.

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1Ah i see  you are squaring both sides Right!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3\[\begin{align} 2\csc^2 x  \frac{1}{2} \csc^2 \frac{x}{2}&=\frac{1}{2}\left(\dfrac{4}{\sin^2x}\frac{1}{\sin^2\frac{x}{2}}\right)\\~\\ &=\frac{1}{2}\left(\dfrac{4}{(\sin x)^2}\frac{1}{\sin^2\frac{x}{2}}\right)\\~\\ &=\frac{1}{2}\left(\dfrac{4}{(2\sin\frac{x}{2}\cos\frac{x}{2})^2}\frac{1}{\sin^2\frac{x}{2}}\right)~~\color{grey}{\because \sin(2x)=2\sin x \cos x}\\~\\ &=\frac{1}{2}\left(\dfrac{4}{4\sin^2\frac{x}{2}\cos^2\frac{x}{2}}\frac{1}{\sin^2\frac{x}{2}}\right)\\~\\ &=\frac{1}{2}\left(\dfrac{1\cos^2\frac{x}{2}}{\sin^2\frac{x}{2}\cos^2\frac{x}{2}}\right)\\~\\ &=\frac{1}{2}\left(\dfrac{1}{\cos^2\frac{x}{2}}\right)\\~\\ \end{align}\] we can take the limit now

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3np:) 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%5E2x1%2F2csc%5E2%28x%2F2%29
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