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Math Help - limit trigo function

  1. #1
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    limit trigo function

    quite rusty...

    Would anyone help with the following, please:

    limc_pi/2[(X+tanC)/(1-XtanC)] =?

    I don't understand why, in a book, they continue thus:

    limc_pi/2[sec2C/(-Xsec2C] = -1/X

    Thank you in advance
    .
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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: limit trigo function

    Your book is using L'H˘pital's rule, which essentially states that if given:

    L=\lim_{u\to c}\frac{f(u)}{g(u)}

    and:

    \lim_{u\to c}f(u)=\lim_{u\to c}g(u)=0 or \lim_{u\to c}f(u)=\lim_{u\to c}g(u)=\infty

    then:

    L=\lim_{u\to c}\frac{f'(u)}{g'(u)}

    Do you see that if you write the limit you are given as:

    -\lim_{c\to\frac{\pi}{2}}\frac{x+\tan(x)}{x\tan(c)-1}

    you have the second indeterminate form above \frac{\infty}{\infty}?
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  3. #3
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    Re: limit trigo function

    Thank you, Mark.
    the
    The indetermination was readily obvious to me.
    My confusion relates to the transition into the sec^2 expressions.

    Would you please comment on that?

    Thanks.
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: limit trigo function

    They are a result of:

    \frac{d}{du}(\tan(u))=\sec^2(u).
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  5. #5
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    Re: limit trigo function

    Of course!

    Thank you very much, Mark!
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