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Math Help - Trigonometric and logarithmic limit?

  1. #1
    Super Member fardeen_gen's Avatar
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    Trigonometric and logarithmic limit?

    Evaluate:
    \lim_{x\rightarrow \infty} x^2\sin\left(\log \sqrt{\cos \frac{\pi}{x}}\right)


    Answer:
    Spoiler:
    -\frac{\pi^2}{4}


    I am unable to get the required answer. How to do it?
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  2. #2
    Senior Member pankaj's Avatar
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    \lim_{x\rightarrow \infty} x^2\sin\left(\log \sqrt{\cos \frac{\pi}{x}}\right)


     <br />
=\lim_{x\rightarrow \infty} x^2\frac{\sin\left(\log \sqrt{\cos \frac{\pi}{x}}\right)}{log\sqrt{\cos\frac{\pi}{x}}  }<br />
log\sqrt{\cos\frac{\pi}{x}}

    =-\pi^2\lim_{x\rightarrow \infty} \frac{\sin\left(\log \sqrt{\cos \frac{\pi}{x}}\right)}{log\sqrt{\cos\frac{\pi}{x}}  }<br />
. \frac{log(1+\sqrt{\cos\frac{\pi}{x}}-1)}{\sqrt{cos\frac{\pi}{x}}-1}. \frac{1-\sqrt{cos\frac{\pi}{x}}}{\frac{\pi^2}{x^2}}

    =-\pi^2\lim_{x\rightarrow \infty} \frac{\sin\left(\log \sqrt{\cos \frac{\pi}{x}}\right)}{log\sqrt{\cos\frac{\pi}{x}}  }<br />
. \frac{log(1+\sqrt{\cos\frac{\pi}{x}}-1)}{\sqrt{cos\frac{\pi}{x}}-1}. \frac{1-\cos\frac{\pi}{x}}{\frac{\pi^2}{x^2}}. \frac{1}{1+\cos\frac{\pi}{x}}

     <br />
=-\frac{\pi^2}{4}<br />
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  3. #3
    Senior Member Spec's Avatar
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    A little explanation of the last part:

    \lim_{x\to \infty}\frac{1-\cos\frac{\pi}{x}}{\frac{\pi^2}{x^2}}=\lim_{t\to 0}\frac{1-\cos t}{t^2}=\lim_{t\to 0}\frac{1-\cos t}{t^2}\frac{1+\cos t}{1+\cos t} =\lim_{t\to 0}\frac{\sin^2 t}{t^2(1+\cos t)}=\lim_{t\to 0}\frac{\sin t}{t}\frac{\sin t}{t}\frac{1}{1+\cos t}=\frac{1}{2}
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  4. #4
    Super Member Random Variable's Avatar
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    Does  \lim_{x\rightarrow \infty} \frac{log(1+\sqrt{\cos\frac{\pi}{x}}-1)}{\sqrt{cos\frac{\pi}{x}}-1} = 1?

    EDIT: I take that back. It's indeterminate in that form. What am I missing?
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  5. #5
    Senior Member Spec's Avatar
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    Make the substitution t=\sqrt{\cos {\frac{\pi}{x}}}-1 and you get t\to 0 as x\to \infty.

    Standard limit: \lim_{x\to 0}\frac{\ln(1+x)}{x}=1
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