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Math Help - trig. limit

  1. #1
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    trig. limit

    lim (x -> 0) sin(x)/(x+tan(x))

    Can someone show me step by step method of solving this. I tried it, but my answer doesn't match the one in the book.
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  2. #2
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    \lim_{x\to\rightarrow{0}}\frac{sin(x)}{x+tan(x)}

    Divide top and bottom by x:

    \lim_{x\to\rightarrow{0}}\frac{\frac{sin(x)}{x}}{1  +\frac{tan(x)}{x}}

    You can see the familiar limits in the num. and den. See what it is now?.
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by cityismine View Post
    lim (x -> 0) sin(x)/(x+tan(x))

    Can someone show me step by step method of solving this. I tried it, but my answer doesn't match the one in the book.
    It is commonly known that \tan(x)\sim{x} and \sin(x)\sim{x}

    \therefore\lim_{x\to{0}}\frac{\sin(x)}{x+\tan(x)}=  \lim_{x\to{0}}\frac{x}{2x}=\frac{1}{2}

    Alternatively

    you can divide by x to get

    \lim_{x\to{0}}\frac{\frac{\sin(x)}{x}}{1+\frac{\ta  n(x)}{x}}

    Evaluating each separately we get

    \lim_{x\to{0}}\frac{\sin(x)}{x}=\lim_{x\to{0}}\fra  c{\sin(x)-\sin(0)}{x-0}=\bigg(\sin(x)\bigg)'\bigg|_{x=0}=1

    and \lim_{x\to{0}}\frac{\tan(x)}{x}=\lim_{x\to{0}}\fra  c{\tan(x)-\tan(0)}{x-0}=\bigg(\tan(x)\bigg)'\bigg|_{x=0}=1

    Thus we have

    \frac{1}{1+1}=\frac{1}{2}

    also you could utilize power series but that might be too advanced
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  4. #4
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    The division by x method works out perfectly, thanks!
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