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Math Help - Limits (URGENT)

  1. #1
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    Limits (URGENT)

    lim x->ln2 of (e^2x - 4) / (e^x - 2)

    when ln2 is plugged in, there is an answer of 0/0 which is not what im looking for so i need to somehow factor it or something


    lim x->6 of [ln(x-5)] / [ln(x-5)^3]

    when 6 is plugged in, there is an answer of 0/0 which is again not what im looking for so i need to somehow factor it


    PLEASE HELP! THANKS
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  2. #2
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    \lim_{x \to \ln 2} \frac{e^{2x} - 4}{e^{x} - 2} = \lim_{x \to \ln 2} \frac{\left(e^{x}\right)^2 - 4}{e^x - 2}

    Apply difference of squares to the numerator.

    \lim_{x \to 6} \frac{\ln (x - 5)}{\ln (x - 5)^3}

    Recall that \ln a^b = b\ln a. Bring down the power and something should cancel
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  3. #3
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    Thanks!
    i figured out the first one right away after you said about the difference of squares
    and i brought down the 3 from the second question but im still lost from there. do i change ln to the loge? because i cant plug in 6 yet as ill still get 0/0
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  4. #4
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    \frac{\ln(x-5)}{\ln(x-5)^3} = \frac{\ln(x-5)}{3\ln(x-5)}

    What cancels?
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  5. #5
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    yea i saw that hahah, but like, whats the point of the 6?
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