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Math Help - L'Hopitals Rule

  1. #1
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    L'Hopitals Rule

    I have these 3 questions, and I think I need to use L'Hopitals rule. But I Have no idea how to go about that. Anything I've read about it confuses me.

    a)
    lim x->1 (ln x)/(x-1)

    b) lim x-> infinity x/ln(1+2e^x)

    c) lim x->0 (cot2x)(sin6x)

    Please help
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Ok

    Quote Originally Posted by dangerous_dave View Post
    I have these 3 questions, and I think I need to use L'Hopitals rule. But I Have no idea how to go about that. Anything I've read about it confuses me.

    a)
    lim x->1 (ln x)/(x-1)

    b) lim x-> infinity x/ln(1+2e^x)

    c) lim x->0 (cot2x)(sin6x)

    Please help
    L'hopital's is the EASIEST way to do this....i will do the hardest one for you \lim_{x \to {\infty}}\frac{x}{ln\bigg(1+2e^{x}\bigg)}\Rightarr  ow{\lim_{x \to {\infty}}\frac{1}{\frac{2e^{x}}{1+2e^{x}}}}\Righta  rrow{1}
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  3. #3
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    Could you please explain how L'Hopitals works? I don't understand at all...
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  4. #4
    Super Member angel.white's Avatar
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    Quote Originally Posted by dangerous_dave View Post
    Could you please explain how L'Hopitals works? I don't understand at all...
    If you have a limit which takes the form of \pm \frac \infty \infty or  \frac 00 then the limit will be equal to the derivative of the numerator over the derivative of the denominator.

    For example \lim_{x\to 0} \frac {sin~x}{x} takes the form of 0/0 so we apply L'Hospital's rule:

    \lim_{x\to 0} \frac {sin~x}{x} ~~~~=~~~~ \lim_{x\to 0} \frac {\frac d{dx} sin~x}{\frac {dx}{dx}} ~~~~=~~~~ \lim_{x\to 0} \frac {cos~x}{1}~~~~=~~~~cos~0~~~~=~~~~1
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  5. #5
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    Wow thats a whole lot easier than other sites like to make out. Thanks heaps
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  6. #6
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    Does "Dangerous Dave" come from the 1990 computer game?

    Quote Originally Posted by dangerous_dave View Post
    a)
    lim x->1 (ln x)/(x-1)
    This is nothing to do with L'Hopital rule. This is the definition of derivative of \ln x at x=1.

    c) lim x->0 (cot2x)(sin6x)
    \cot 2x \sin 6x = \cos 2x \left( \frac{\sin 6x}{\sin 2x} \right).
    Note, \cos 2x \to 1.
    While, \frac{\sin 6x}{\sin 2x} = \frac{\sin 6x}{6x} \cdot \frac{2x}{\sin 2x}\cdot 3\to 1\cdot 1\cdot 3 = 3
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