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Math Help - Getting the limit

  1. #1
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    Getting the limit

    Reviewed my textbook on limits. It went over various types including the difference between limits that go to infinity. A problem occurred to me that the book didn't cover so I'm posting this as a puzzle.

    Let's define this function: g(x) = exp(x) - ln(x) and let x approach infinity. What value does g(x) take on when x approaches infinity? (I'm wondering whether using the series approach will solve this problem).
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  2. #2
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    y=e^x,\;\;\;x=log_ey

    Hence y=log_ex is the mirror-image of y=e^x through the line y=x

    The functions differ by ever-increasing magnitude
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  3. #3
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    Using a "series approach":

    e^x \ge 1+x+\frac{x^2}{2} and \ln x \le x+1 (for x\ge 1) we get

    e^x-\ln x \ge (1+x+\frac{x^2}{2}) - (x+1) = \frac{x^2}{2}

    from which the limit follows, since \lim_{x\rightarrow\infty} \frac{x^2}{2} = \infty

    Another approach:

    \lim_{x\rightarrow\infty} e^x-\ln x = \lim_{x\rightarrow\infty} (1-\frac{\ln x}{e^x}) e^x

    and since \lim_{x\rightarrow\infty} 1-\frac{\ln x}{e^x} = 1 and

    \lim_{x\rightarrow\infty} e^x = \infty we get the desired result.
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  4. #4
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    Quote Originally Posted by wonderboy1953 View Post
    Reviewed my textbook on limits. It went over various types including the difference between limits that go to infinity. A problem occurred to me that the book didn't cover so I'm posting this as a puzzle.

    Let's define this function: g(x) = exp(x) - ln(x) and let x approach infinity. What value does g(x) take on when x approaches infinity? (I'm wondering whether using the series approach will solve this problem).
    Since exp(x) grows faster than any power of x, and log(x) grows more slowly than any non-zero power of x their difference grows faster than any power of x.

    CB
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  5. #5
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    Quote Originally Posted by CaptainBlack View Post
    Since exp(x) grows faster than any power of x, and log(x) grows more slowly than any non-zero power of x their difference grows faster than any power of x.

    CB
    That's my line of reasoning CB. I was just looking for a more rigorous line of thought on what I suspected.
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  6. #6
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    Quote Originally Posted by wonderboy1953 View Post
    That's my line of reasoning CB. I was just looking for a more rigorous line of thought on what I suspected.
    It is actually rigorous, I was just expressing known big-O results in colloquial English.

    CB
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  7. #7
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    Also,

    e^x>2x^2lnx

    \displaystyle\lim_{x \to \infty}\left(e^x-lnx\right)\;>\;\lim_{x \to \infty}\left(2x^2-1\right)lnx

    lnx>1 for x>e
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