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Thread: 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 $\displaystyle \lim_{x \to {\infty}}\frac{x}{ln\bigg(1+2e^{x}\bigg)}\Rightarr ow{\lim_{x \to {\infty}}\frac{1}{\frac{2e^{x}}{1+2e^{x}}}}\Righta rrow{1}$
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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 $\displaystyle \pm \frac \infty \infty$ or $\displaystyle \frac 00$ then the limit will be equal to the derivative of the numerator over the derivative of the denominator.

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

    $\displaystyle \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 $\displaystyle \ln x$ at $\displaystyle x=1$.

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