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  1. #1
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    limit problem_ass4_9

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

    How I can solve this ? I tried L'Hospital's but I stuck ...
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  2. #2
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    Quote Originally Posted by gilyos View Post
    lim_{x \to \infty} \frac{ln(e^{3x}-5x)}{x}

    How I can solve this ? I tried L'Hospital's but I stuck ...
    hmmm.... after using L.H rule once you should get

    \lim_{x \to \infty}\frac{\frac{3e^{3x}-5}{e^{3x}-5x}}{1}= \lim_{x \to \infty}\frac{3e^{3x}-5}{e^{3x}-5x}

    This is still of the form infinity over infinity.

    use L.H's Rule again

    =\lim_{x \to \infty} \frac{9e^{3x}}{3e^{3x}-5}

    Finally use it one more time and use algebra to simplify.

    From a different point of view note that

    for large x \ln(e^{3x}-5x) \approx \ln(e^{3x})= 3x

    so the limit is 3
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  3. #3
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     \lim_{x \to \infty}\frac{3e^{3x}-5}{e^{3x}-5x}= \frac{[\infty-5]}{[\infty]}

    There permission here to L'Hospital's ??
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  4. #4
    Member Abu-Khalil's Avatar
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    \lim_{x\to\infty}\frac{3e^{3x}-5}{e^{3x}-5x}=\lim_{x\to\infty}\frac{3-\frac{5}{e^{3x}}}{1-\frac{5x}{e^{3x}}}=3

    Also,
    \lim_{x\to\infty}\frac{\log\left(e^{3x}-5x\right)}{x}=\log\left\{\lim_{x\to\infty}\left(e^  {3x}-5x\right)^\frac{1}{x}\right\} and \underbrace{\left(e^{3x}-\frac{1}{2}e^{3x}\right)^{\frac{1}{x}}}_{\to e^3}\leq\left(e^{3x}-5x\right)^{\frac{1}{x}}\leq \underbrace{\left(e^{3x}+e^{3x}\right)^{\frac{1}{x  }}}_{\to e^3}
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  5. #5
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    Quote Originally Posted by gilyos View Post
    lim_{x \to \infty} \frac{ln(e^{3x}-5x)}{x}

    How I can solve this ? I tried L'Hospital's but I stuck ...
    For very large x, e^{3x} is much larger than 5x so this has the same limit as \frac{ln(e^{3x})}{x}= \frac{3x}{x}= 3.
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by gilyos View Post
    lim_{x \to \infty} \frac{ln(e^{3x}-5x)}{x}

    How I can solve this ? I tried L'Hospital's but I stuck ...
    Exactly what HallsOfIvy said but a little bit more formal \ln\left(e^{3x}-5x\right)=3x+\ln\left(1-\tfrac{5x}{e^{3x}}\right) and since \frac{5x}{e^{3x}}\underset{x\to\infty}\to 0 we may conclude that \ln\left(1-\tfrac{5x}{e^{3x}}\right)\underset{x\to\infty}{\si  m}\frac{-5x}{e^{3x}} and the conclusion easily follows.
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