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Math Help - Need help with this limit!

  1. #1
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    Smile Need help with this limit!

     \lim_{x\to0^+}(9x+1)^{cotx}

    Hopefull I am at least on the right track, here's what I tried to do:

    Let y=(9x+1)^{cotx}

    \lim_{x\to0^+}lny=cot(x)ln(9x+1)

    \lim_{x\to0^+}lny=\frac{ln(9x+1)}{\frac{1}{cot(x)}  }

    \lim_{x\to0^+}lny=\frac{\frac{9}{9x+1}}{\frac{csc^  2x}{cot^2x}}

    And I don't know what to do next.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by xxlvh View Post
     \lim_{x\to0^+}(9x+1)^{cotx}

    Hopefull I am at least on the right track, here's what I tried to do:

    Let y=(9x+1)^{cotx}

    \lim_{x\to0^+}lny=cot(x)ln(9x+1)

    \lim_{x\to0^+}lny=\frac{ln(9x+1)}{\frac{1}{cot(x)}  }

    \lim_{x\to0^+}lny=\frac{\frac{9}{9x+1}}{\frac{csc^  2x}{cot^2x}}

    And I don't know what to do next.
    note that \frac 1{\cot x} = \tan x. The derivative is \sec^2 x (if you simplified your denominator, you would have gotten the same thing). Now just plug in x = 0 and continue

    (You should have the \lim_{x \to 0^+} on the right side of the equation also)
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