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Math Help - trigonometric limit

  1. #1
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    trigonometric limit

    \lim_{x \to 0}\frac{x-tan2x}{sin2x}
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  2. #2
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    Quote Originally Posted by FLTR View Post
    \lim_{x \to 0}\frac{x-tan2x}{sin2x}
    The function,
    f(x)=\frac{x-\tan 2x}{\sin 2x}
    Can be expressed equivalently as,
    f(x)=\frac{x}{\sin 2x}-\frac{\tan 2x}{\sin 2x}
    If we "cancel" to get,
    g(x)=\frac{x}{\sin 2x}-\sec 2x
    Then the functions, f(x),g(x) will agree on some open interval containing 0 except possiblly at x=0 itself.
    Thus, the limit exististence and value is equivalent to,
    \lim_{x\to 0} \frac{x}{\sin 2x} - \sec 2x
    Express as,
    \lim_{x\to 0} \frac{1}{2}\cdot \frac{2x}{\sin 2x} - \sec 2x
    We note that,
    \lim_{x\to 0} \frac{2x}{\sin 2x}=1 by the limit composition rule and one of the famous limits.
    Thus, since the functional limits exists so does its sum which is,
    \frac{1}{2} \lim_{x\to 0}\frac{2x}{\sin 2x} - \lim_{x\to 0}\sec 2x=\frac{1}{2}-1=-\frac{1}{2}
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  3. #3
    TD!
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    Taylor works fast here too, sin(a) =~ tan(a) =~ a for a arround 0, so:

    <br />
\mathop {\lim }\limits_{x \to 0} \frac{{x - \tan 2x}}{{\sin 2x}} = \mathop {\lim }\limits_{x \to 0} \frac{{x - 2x}}{{2x}} = \mathop {\lim }\limits_{x \to 0} \frac{{ - x}}{{2x}} =  - \frac{1}{2}<br />
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