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anonymous

  • 5 years ago

convergence test (|cos x|)/(x^2-|sin x|) dx limits 1 to infinity

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  1. anonymous
    • 5 years ago
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    limits 1 to infinity converge or diverge and proof (|cos x|)/(x^2 - |sin x|)

  2. anonymous
    • 5 years ago
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    The absolute value of cosine is always less than or equal to one. So you may take,\[\frac{|\cos x|}{x^2 - |\sin x|} \le \frac{1}{x^2- |\sin x|}=\frac{1/x^2}{1-\frac{|\sin x|}{x^2}}\]

  3. anonymous
    • 5 years ago
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    \[\lim_{x \rightarrow \infty}\frac{|\cos x|}{x^2 - |\sin x|} \le \lim_{x \rightarrow \infty}\frac{1/x^2}{1-\frac{|\sin x|}{x^2}}\]\[=\frac{\lim_{x \rightarrow \infty}1/x^2}{\lim_{x \rightarrow \infty}1-\lim_{x \rightarrow \infty}\frac{|\sin x|}{x^2}}=\frac{0}{1-0}=0\]

  4. anonymous
    • 5 years ago
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    Damn, just spotted the 'dx'. I thought you wanted to see what the limit of this thing was. Are you using the integral test?

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