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Math Help - differentiation of 1n (3x + e^x)

  1. #1
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    differentiation of 1n (3x + e^x)

    Hi

    So I am applying l'Hospital's Rule to

    lim (9x)/(ln (3x + e^x))
    x-> infinity

    What I wondering is ln (3x + e^x) = ln 3x + ln e^x and then do they both just become 1/3x and 1/e^x?

    Any insight would be appreciated
    Thanks

    Still a calculus beginner
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by calcbeg View Post
    Hi

    So I am applying l'Hospital's Rule to

    lim (9x)/(ln (3x + e^x))
    x-> infinity

    What I wondering is ln (3x + e^x) = ln 3x + ln e^x and then do they both just become 1/3x and 1/e^x?

    Any insight would be appreciated
    Thanks

    Still a calculus beginner
    ln (3x + e^x) \neq ln 3x + ln e^x
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  3. #3
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    okay so how do I solve ln (3x +e^x)? what is the derivative?

    Calculus beginner
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  4. #4
    Member WhoCares357's Avatar
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    Quote Originally Posted by calcbeg View Post
    okay so how do I solve ln (3x +e^x)? what is the derivative?

    Calculus beginner
    \frac{d}{dx}(ln(u))=\frac{du}{u}
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  5. #5
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    Quote Originally Posted by calcbeg View Post
    okay so how do I solve ln (3x +e^x)? what is the derivative?

    Calculus beginner
    Using the chain rule:

    Let 3x + e^x = u

    Then ln(3x+e^x) = ln(u)

    Now: \frac{d}{dx}ln(u) = \frac{d}{du}ln(u) \times \frac{du}{dx} = \frac{1}{u} \times (3 + e^x) = \frac{3 + e^x}{3x + e^x}
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  6. #6
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    So lim 9x/ ln (3x + e^x) becomes 9/ (3 + e^x)/(3x + e^x)
    x->infinity

    which becomes (27x + 9e^x)/(3 + e^x) which is again infinity/infinity

    so I am thinking I need to apply the l'Hospital's rule again

    so now I apply the quotient rule so it becomes

    ((27 + 9e^x)(e^x) - (e^x)(27x + 9e^x))/((e^x)^2)

    so (27 + 9e^x - 27x-9e^x)/e^x = (27 - 27x)/e^x so I still have infinity/infinity

    where am I going wrong????

    Thanks

    Calculus Beginner
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  7. #7
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    Sorry I am now confusing myself

    Am I right.... applying l'Hospital's rule a second time I get

    lim (27x + 9e^x)/ (3+ e^x) = lim 27 + 9e^x/e^x
    x-> infinity

    so then I apply l'hospital's rule a third time to get

    lim 9e^x/e^x = 9
    x-> infinity

    Am I right???

    Thanks

    Calculus Beginner
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  8. #8
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    You have to break it up:

    \lim_{x \to \infty} \frac{27x + 9e^x}{3 + e^x} = \lim_{x \to \infty} \frac{\frac{27x}{x}}{\frac{3}{x} + \frac{e^x}{x}} + \frac{\frac{9e^x}{e^x}}{\frac{3}{e^x} + \frac{e^x}{e^x}}

    You'll find that the first part has a limit of 0 and the second part has a limit of 9. So the overall limit is 9.
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  9. #9
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    Quote Originally Posted by calcbeg View Post
    Sorry I am now confusing myself

    Am I right.... applying l'Hospital's rule a second time I get

    lim (27x + 9e^x)/ (3+ e^x) = lim 27 + 9e^x/e^x
    x-> infinity

    so then I apply l'hospital's rule a third time to get

    lim 9e^x/e^x = 9
    x-> infinity

    Am I right???

    Thanks

    Calculus Beginner
    Yes, this is correct, however there really isn't any need to abuse L'Hospital's so much!

    After using it for the first time, simply divide each term by e^x and get:

    \lim_{x \to \infty} \frac{27x + 9e^x}{3 + e^x} = \lim_{x \to \infty} \frac{\frac{27x}{e^x} + \frac{9e^x}{e^x}}{\frac{3}{e^x} + \frac{e^x}{e^x}} = \frac{0 + 9}{0 + 1} = 9
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