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Math Help - Trig Limits

  1. #1
    Newbie mushufasa11's Avatar
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    Trig Limits

    1.) limx-->0; (x + sinx) / (x + cosx)

    2.) limx-->infinite; (e^-x)cosx

    3. limx-->0; (xcscx + 1) / (xcscx)


    thanks
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  2. #2
    Junior Member
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    not sure but

    1.) limx-->0; (x + sinx) / (x + cosx)
    dividing the top and bottom by x
    =limx-->0; [1 + (sinx)/x] / [1 + (cosx)/x]
    limx-->0 (sinx)/x=1 ;limx-->0 (cosx)/x=0
    then continue... u should get 2

    2.) limx-->infinite; (e^-x)cosx
    =limx-->infinite; (cosx)/e^x
    i think it is zero since limx-->infinite; e^x=infinite, but might also be undefined if x=n pi/2

    3.) limx-->0; (xcscx + 1) / (xcscx)
    divide top and bottom by xcsc+1
    =limx-->0; [1+ (sinx)/x] / [(xcscx)/(xcscx)]
    =limx-->0; [1+ (sinx)/x]
    limx-->0 (sinx)/x=1; u should get 2
    Last edited by calculus_jy; September 17th 2007 at 01:35 AM. Reason: typo
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  3. #3
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    Hello, mushufasa11!

    An alternate method for #3.

    It uses the theorem: . \lim_{\theta\to0}\frac{\sin\theta}{\theta} \:=\:1


    3)\;\;\lim_{x\to0}\frac{x\csc x + 1}{x\csc x}

    Multiply top and bottom by \sin x

    Then we have: . \lim_{x\to0}\frac{x + \sin x}{x} \;=\;\lim_{x\to0}\left(1 + \frac{\sin x}{x}\right) \;=\;1 + 1 \;=\;2

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