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Math Help - limit as h approaches 0

  1. #1
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    limit as h approaches 0

    What is the

    limit of [ tan(pi/6+h) - tan(pi/6) ] / h

    as h approaches 0


    A) sqrt(3) / 3
    B) 4/3
    C) sqrt(3)
    D) 0
    E. 3/4
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  2. #2
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    Quote Originally Posted by DINOCALC09 View Post
    What is the

    limit of [ tan(pi/6+h) - tan(pi/6) ] / h

    as h approaches 0
    First, \tan (x+y) = \frac{\tan x + \tan y}{1-\tan x \tan y}.

    This means, \tan \left( \frac{\pi}{6} + h \right) = \frac{\tan \frac{\pi}{6} + \tan h}{1 - \tan \frac{\pi}{6} \cdot \tan h} = \frac{\frac1{\sqrt{3}} + \tan h}{1 - \tan h \cdot \frac1{\sqrt{3}}}

    Thus, \tan \left( \frac{\pi}{6} + h \right) - \tan \frac{\pi}{6} = \frac{\frac1{\sqrt{3}} + \tan h}{1 - \tan h \cdot \frac1{\sqrt{3}}} - \frac1{\sqrt{3}} = \frac{\frac 43\tan h}{1 - \tan h \cdot \sqrt{3}}

    If you divide this by h, you get, \frac{\frac43\frac{\tan h}{h}}{1 - \tan h \cdot \sqrt{3}}.

    Take h\to 0 and use the fact that \lim_{h\to 0}\frac{\tan h}{h} = 1 and we get that the limit is \frac{4}{3}.
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  3. #3
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    One might note that:
    \lim _{h \to 0} \left[ {\frac{{\tan \left( {\frac{\pi }{6} + h} \right) - \tan \left( {\frac{\pi }{6}} \right)}}{h}} \right] = \sec ^2 \left( {\frac{\pi }{6}} \right)
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  4. #4
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    Quote Originally Posted by Plato View Post
    One might note that:
    \lim _{h \to 0} \left[ {\frac{{\tan \left( {\frac{\pi }{6} + h} \right) - \tan \left( {\frac{\pi }{6}} \right)}}{h}} \right] = \sec ^2 \left( {\frac{\pi }{6}} \right)
    It did not seem to me that he knew what a derivative is.
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  5. #5
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    Quote Originally Posted by ThePerfectHacker View Post
    It did not seem to me that he knew what a derivative is.
    I bet you that is what the question is about.
    It is just too perfectly setup.
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