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Math Help - Question on Limits

  1. #1
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    Question on Limits

    Evaluate \lim_{x \to \infty} \frac{(\frac{10}{11})^x}{(\frac{9}{10})^x + (\frac{11}{12})^x}

    Instinctively it's \frac{1}{2}, just that I'm not too sure how to go about showing it.
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  2. #2
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    Re: Limit

    Note that 9/10, 10/11, and 11/12 are all less than one, so \left(\frac{9}{10}\right)^x, \left(\frac{10}{11}\right)^x, and \left(\frac{11}{12}\right)^x all tend to zero as x goes to infinity. The largest of these numbers is 11/12, so it goes to zero most slowly. Thus, we multiply both the numerator and denominator by \left(\frac{12}{11}\right)^x:
    \lim_{x\to\infty}\frac{(\frac{10}{11})^x}{(\frac{9  }{10})^x+(\frac{11}{12})^x}=\lim_{x\to\infty}\frac  {(\frac{10}{11})^x(\frac{12}{11})^x}{((\frac{9}{10  })^x+(\frac{11}{12})^x)(\frac{12}{11})^x}
    =\lim_{x\to\infty}\frac{(\frac{120}{121})^x}{(\fra  c{54}{55})^x+1}.
    Note that 120/121 and 54/55 are both less than one, as expected, and so as x goes to infinity, the numerator of the above goes to zero, and the denominator goes to one. Thus the limit is 0.

    --Kevin C.
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  3. #3
    Super Member wingless's Avatar
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    \lim_{x\to\infty} \frac{\left ( \frac{10}{11} \right )^x}{\left ( \frac{9}{10} \right )^x + \left ( \frac{11}{12} \right )^x}

    The general approach for these limits is factoring the largest term of the denominator. It is \left ( \frac{11}{12} \right )^x here.

    \lim_{x\to\infty} \frac{\left ( \frac{11}{12} \right )^x \cdot \left ( \frac{120}{121} \right )^x}{\left ( \frac{11}{12} \right )^x \cdot \left [ 1 + \left ( \frac{108}{110} \right )^x \right ] }

    \lim_{x\to\infty} \frac{\left ( \frac{120}{121} \right )^x}{1 + \left ( \frac{108}{110} \right )^x }

    We know that \lim_{x\to\infty} r^x = 0 for |r|<1. So those fractions will go to zero.

    \lim_{x\to\infty} \frac{0}{1+0}=0
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  4. #4
    Super Member flyingsquirrel's Avatar
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    Hello
    Quote Originally Posted by h2osprey View Post
    Evaluate \lim_{x \to \infty} \frac{(\frac{10}{11})^x}{(\frac{9}{10})^x + (\frac{11}{12})^x}

    Instinctively it's \frac{1}{2}, just that I'm not too sure how to go about showing it.
    Another possible approach : 0<\frac{(\frac{10}{11})^x}{(\frac{9}{10})^x + (\frac{11}{12})^x}\leq\frac{(\frac{10}{11})^x}{(\f  rac{11}{12})^x}=\left(\frac{120}{121}\right)^x\und  erset{x\to\infty}{\to}0 and we can conclude using the squeezing theorem.
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