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Math Help - L'Hopital's limit

  1. #1
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    L'Hopital's limit

    \lim_{x \to \infty} (4^x + 7^x)^(1/x)

    First I took the ln of (4^x + 7^x)^(1/x) to get (4^x + 7^x)/x and derived the top and bottom, so I got a function of
    (4^x *ln4 + 7^x *ln7)/(4^x + 7^x)

    and I took the derivative of the top and bottom again, but I could tell already that things were going nowhere :/
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  2. #2
    A Plied Mathematician
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    Hmm. Let's see.

    \ln\left(\lim_{x\to\infty}(4^{x}+7^{x})^{1/x}\right)=\lim_{x\to\infty}\ln\left((4^{x}+7^{x})^  {1/x}\right)=\lim_{x\to\infty}\left[\frac{\ln(4^{x}+7^{x})}{x}\right]
    =\underbrace{\lim_{x\to\infty}\left[\frac{4^{x}\ln(4)+7^{x}\ln(7)}{4^{x}+7^{x}}\:\frac  {1}{1}\right]}_{\text{L'Hopital's Rule}}.

    There is where you ended. At this point I would divide top and bottom by 7^{x}, instead of using L'Hopital's Rule again.
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  3. #3
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    ohhhhhhhh i see now. Dividing the top and bottom by 7^x yields \lim_{x \to \infty} ((((4^x)*ln4) /7^x) + ln7)/((4^x/7^x)+1) and 4^x*ln4/7^x and 4^x / 7^x both go to zero since 7^x will grow at a faster rate than 4^x, so ln \lim_{x \to \infty} (4^x + 7^x)^(1/x) = e^(ln7) = 7

    Thank you so much!

    But real quick: Why did you have to divide by 7^x to get the right answer?
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  4. #4
    A Plied Mathematician
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    You got it, and you're welcome!
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