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Math Help - 3 part Question -- main problem with limit of arctan

  1. #1
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    3 part Question -- main problem with limit of arctan

    I believe I found the answer to part a), and I also think I could find part c).

    I just don't know how to evaluate part b). I was thinking of using the squeeze theorem (or whatever you'd like to call it), but I don't know...

    {-\pi \over 2} \leq tan^-1({-16 \over x-2}) \leq {\pi \over 2}

    I don't really know what to do with this
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  2. #2
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    Re: 3 part Question -- main problem with limit of arctan

    as x \to 2^+ , \frac{-16}{x-2} \to -\infty

    now check out the graph of y = \arctan{x} ... what value does y approach as x \to -\infty ?
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  3. #3
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    Re: 3 part Question -- main problem with limit of arctan

    It is approaching -\pi \over 2
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  4. #4
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    Re: 3 part Question -- main problem with limit of arctan

    Quote Originally Posted by Robinator View Post
    It is approaching -\pi \over 2
    therefore, \lim_{x \to 2^+} \arctan\left(\frac{-16}{x-2}\right) = \, ?
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  5. #5
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    Re: 3 part Question -- main problem with limit of arctan

    -\pi \over 2 Lol. Thank you.
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