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Math Help - LIMxgoesto0(tan(7x)cosec(3x))

  1. #1
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    LIMxgoesto0(tan(7x)cosec(3x))

    any plan of attack would be great.
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by Nicholas View Post
    any plan of attack would be great.
    If you know that \sin(ax)=ax+o(x) and \cos(ax)= 1-ax+o(x) for x\to 0, you are almost done:

    \lim_{x\to 0}\left(\tan(7x)\cdot \mathrm{cosec}(3x)\right)=\lim_{x\to0}\frac{\sin(7  x)}{\cos(7x)\cdot \sin(3x)}

    =\lim_{x\to 0}\frac{7x+o(x)}{(1-7x+o(x))\cdot (3x+o(x))}=\lim_{x\to 0}\frac{7x}{(1-7x)\cdot 3x}=\ldots

    Where o(x) is Landau's little-oh notation: \lim_{x\to 0}\frac{o(x)}{x}=0.
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  3. #3
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    Hello, Nicholas!

    You're expected to know this theorem:


    . . \lim_{\theta\to0}\frac{\sin\theta}{\theta} \;=\;\lim_{\theta\to0}\frac{\theta}{\sin\theta} \:=\:1


    \lim_{x\to0}\tan7x\csc3x

    We have: . \tan7x\csc3x \:=\:\frac{\sin7x}{\cos7x}\cdot\frac{1}{\sin3x}


    Multiply by \frac{7x}{7x} and \frac{3x}{3x}\!:\quad {\color{blue}\frac{7x}{7x}}\cdot\frac{\sin7x}{\cos  7x} \cdot {\color{blue}\frac{3x}{3x}}\cdot\frac{1}{\sin3x}

    . . =\;\frac{7x}{1}\cdot\frac{\sin7x}{7x}\cdot\frac{1}  {\cos7x} \cdot \frac{1}{3x}\cdot\frac{3x}{\sin3x} \;=\;\frac{7x}{3x}\cdot\frac{\sin7x}{7x}\cdot\frac  {1}{\cos7x} \cdot \frac{3x}{\sin3x}


    Therefore: . \lim_{x\to0}\left(\frac{7}{3}\cdot\frac{\sin7x}{7x  }\cdot\frac{1}{\cos7x}\cdot\frac{3x}{\sin3x}\right  ) \;=\;\frac{7}{3}\cdot 1\cdot\frac{1}{1}\cdot 1 \;=\;\frac{7}{3}


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  4. #4
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    Apologies, I am a first year uni student and have not yet learnt that theorm. Thankyou very much though for explaining it! Much appreciated.
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