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Thread: Trigonometric and logarithmic limit?

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

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


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


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


    $\displaystyle
    =\lim_{x\rightarrow \infty} x^2\frac{\sin\left(\log \sqrt{\cos \frac{\pi}{x}}\right)}{log\sqrt{\cos\frac{\pi}{x}} }
    $$\displaystyle log\sqrt{\cos\frac{\pi}{x}}$

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

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

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

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

    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 $\displaystyle t=\sqrt{\cos {\frac{\pi}{x}}}-1$ and you get $\displaystyle t\to 0$ as $\displaystyle x\to \infty.$

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