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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.

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  1. welshfella
    • one year ago
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    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
    • one year ago
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    i think we can try to simplify the expression using trig identities

  3. ganeshie8
    • one year ago
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    \[\begin{align} 2\csc^2 x - \frac{1}{2} \csc^2 \frac{x}{2}&=\frac{1}{2}\left(\dfrac{4}{\sin^2x}-\frac{1}{\sin^2\frac{x}{2}}\right)\\~\\ &=\frac{1}{2}\left(\dfrac{4}{4\sin^2\frac{x}{2}\cos^2\frac{x}{2}}-\frac{1}{\sin^2\frac{x}{2}}\right)\\~\\ &=\frac{1}{2}\left(\dfrac{1-\cos^2\frac{x}{2}}{\sin^2\frac{x}{2}\cos^2\frac{x}{2}}\right)\\~\\ &=\frac{1}{2}\left(\dfrac{1}{\cos^2\frac{x}{2}}\right)\\~\\ \end{align}\] we can take the limit now

  4. welshfella
    • one year ago
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    yea

  5. welshfella
    • one year ago
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    1/2

  6. thomas5267
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    Ahh that identity looks much better as it avoids the half angles/fractions

  9. welshfella
    • one year ago
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    i didn't know that identity on the second line @ganeshie

  10. welshfella
    • one year ago
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    that half angle one

  11. welshfella
    • one year ago
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    I searched the web to find a half angle identity for sin^ x but couldnt find one

  12. ganeshie8
    • one year ago
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    ohh actually it is just the familiar sin(2x) formula, i skipped several steps in between.. let me add them

  13. thomas5267
    • one year ago
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    \[ \sin^2\theta = \frac{1 - \cos 2\theta}{2} \] Copied shamelessly from Wikipedia.

  14. welshfella
    • one year ago
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    Ah i see - you are squaring both sides Right!

  15. ganeshie8
    • one year ago
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    \[\begin{align} 2\csc^2 x - \frac{1}{2} \csc^2 \frac{x}{2}&=\frac{1}{2}\left(\dfrac{4}{\sin^2x}-\frac{1}{\sin^2\frac{x}{2}}\right)\\~\\ &=\frac{1}{2}\left(\dfrac{4}{(\sin x)^2}-\frac{1}{\sin^2\frac{x}{2}}\right)\\~\\ &=\frac{1}{2}\left(\dfrac{4}{(2\sin\frac{x}{2}\cos\frac{x}{2})^2}-\frac{1}{\sin^2\frac{x}{2}}\right)~~\color{grey}{\because \sin(2x)=2\sin x \cos x}\\~\\ &=\frac{1}{2}\left(\dfrac{4}{4\sin^2\frac{x}{2}\cos^2\frac{x}{2}}-\frac{1}{\sin^2\frac{x}{2}}\right)\\~\\ &=\frac{1}{2}\left(\dfrac{1-\cos^2\frac{x}{2}}{\sin^2\frac{x}{2}\cos^2\frac{x}{2}}\right)\\~\\ &=\frac{1}{2}\left(\dfrac{1}{\cos^2\frac{x}{2}}\right)\\~\\ \end{align}\] we can take the limit now

  16. welshfella
    • one year ago
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    yes I get it now

  17. welshfella
    • one year ago
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    thanks

  18. ganeshie8
    • one year ago
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    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

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