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Math Help - L'Hospitals rule and Mean Value Theorem

  1. #1
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    L'Hospitals rule and Mean Value Theorem

    1) F(x) = \frac{f(1+5x)+f(1+6x)}{x}

    Given f(1) = 0, and f '(1) = 8, evaluate \lim_{x\to0}F(x)

    Limits as it results to \frac{0}{0}, so differentiate top and bottom using Chain Rule:

    \lim_{x\to0} \frac{(5)f'(1+5x)+(6)f'(1+6x)}{1} = 40+48 = 98

    2) If f(2) = 13 and f '(x) ≥ 1 for 4 ≤ x ≤ 4, how small can f(4) possibly be?

    f(4) - f(2) = f'(c)(4-2)
    f(4) = 13 + (2)f'(c) > 1

    ?

    The last step, and where to go from there is what I'm a little shaky on.

    3) \lim_{x\to\infty} (1+\frac{c}{x})^q
    where q = hx
    h and c are both constants, to my knowledge.

    We were taught to do this in a way that seems a little convoluted, but here's as far as I got with this one:

    \lim_{x\to\infty} (1+\frac{c}{x})^q = (e^z)^q = e^g

     z = ln(1+\frac{c}{x}); g = (hx)ln(1+\frac{c}{x})

    From that point we can just take the limit of e to the g'(x), so

    g(x) = (hx)ln(1+\frac{c}{x}) =  \frac{ln(1+\frac{c}{x})}{\frac{1}{hx}}

    Then I get stuck again, because the derivative of the bottom is \frac{-h}{(hx)^2}

    I'm imaging there has to be a way to simplify that, but I can't think of it.
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  2. #2
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    Quote Originally Posted by Open that Hampster! View Post
    1) F(x) = \frac{f(1+5x)+f(1+6x)}{x}

    Given f(1) = 0, and f '(1) = 8, evaluate \lim_{x\to0}F(x)

    Limits as it results to \frac{0}{0}, so differentiate top and bottom using Chain Rule:

    \lim_{x\to0} \frac{(5)f'(1+5x)+(6)f'(1+6x)}{1} = 40+48 = 98


    {\color{red}\mbox{ Of course, you're assuming here that } f'(x) \mbox{ is continuous at } x = 1,} {\color{red}\mbox{ otherwise this is unjustified}}


    2) If f(2) = 13 and f '(x) ≥ 1 for 4 ≤ x ≤ 4, how small can f(4) possibly be?

    f(4) - f(2) = f'(c)(4-2)
    f(4) = 13 + (2)f'(c) > 1

    ?

    The last step, and where to go from there is what I'm a little shaky on.


    {\color{blue}\mbox{ You're almost there:} f(4)=13+2f'(c)>13+2=15}


    3) \lim_{x\to\infty} (1+\frac{c}{x})^q
    where q = hx
    h and c are both constants, to my knowledge.

    We were taught to do this in a way that seems a little convoluted, but here's as far as I got with this one:

    \lim_{x\to\infty} (1+\frac{c}{x})^q = (e^z)^q = e^g


    {\color{red}\mbox{ I don't have the faintest idea what you did here. Imo it is way simpler:}}
    {\color{red}\lim_{x\rightarrow\infty}\left(1+\frac  {c}{x}\right)^q=\left(\lim_{x\rightarrow\infty}\le  ft(1+\frac{c}{x}\right)^x\right)^h=e^{ch}}

    {\color{blue}Tonio}


     z = ln(1+\frac{c}{x}); g = (hx)ln(1+\frac{c}{x})

    From that point we can just take the limit of e to the g'(x), so

    g(x) = (hx)ln(1+\frac{c}{x}) =  \frac{ln(1+\frac{c}{x})}{\frac{1}{hx}}

    Then I get stuck again, because the derivative of the bottom is \frac{-h}{(hx)^2}

    I'm imaging there has to be a way to simplify that, but I can't think of it.
    .
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