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Math Help - limit

  1. #1
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    limit

    hello every body

    i need ahelp

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  2. #2
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    Re: limit

    the limit = 2
    try to find it.....show us some of your work...
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  3. #3
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    Re: limit

    It is easy to get the final answer, but the procedure remain....
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  4. #4
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    Re: limit

    Try L'hospitals rule
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  5. #5
    Senior Member x3bnm's Avatar
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    Re: limit

    We've to find the limit of:

    \lim_{x \to 0}\frac{\sin(\tan(2x))}{\log(x+1)}

    It's in indeterminate form means the function is in the form of \frac{0}{0}.

    So using I'Hopital's rule we get:

    \begin{align*}& \lim_{x \to 0}\frac{\sin{(\tan(2x))}}{\log(x+1)}\\=& \lim_{x \to 0} \frac{2\cos{(\tan(2x))}\sec^{2}(2x)}{\frac{1}{x+1}  }\\=& \lim_{x \to 0} 2\cos{(\tan(2x))}\sec^{2}(2x) (x+1)\\=& (2)(1)(1)(0+1)\\=& 2......\text{[Answer]}\end{align*}

    You can check the answer here:

    \lim_{x \to 0}\frac{\sin(\tan(2x))}{\log(x+1)} - Wolfram|Alpha
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