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Math Help - limit help, please check my working

  1. #1
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    limit help, please check my working

    Use L’Hopitals rule, with a substitution where necessary, to find

     \lim_{x \to 0} \frac{tanx}{tan3x}

    is my method correct,

    differentiate top and bottom,

     \frac{sec^{2}x}{3sec^{2}3x}

     sec^{2} x = tan^{2}x +1

     sec^{2}3x = tan^{2}3x +1

     \frac{tan^{2}x+1}{3tan^{2}3x +3}

    now evaluate the limit and it gives  \frac{1}{3}  ?

    IS this correct?

    Thank you.
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  2. #2
    fkf
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    Re: limit help, please check my working

    Yes! That is correct.
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  3. #3
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    Re: limit help, please check my working

    thank you.
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