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Math Help - Need help on L'Hospitals, Partial fractions

  1. #1
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    Need help on L'Hospitals, Partial fractions

    Hi guys I have a test tomorrow and am confused on a few homework questions on L'Hospital's Rule and partial fractions

    1. Using L'Hospital's Rule:
    \lim_{x\rightarrow{0}}\frac{e^x-1-x}{x^2}

    2. L'Hospital's Rule:
    \lim_{x\rightarrow{{\pi/2}^+}}\frac{cosx}{1-sinx}

    3. Partial Fractions:
    \int \frac{4y^2-7y-12}{y(y+2)(y-3)}dy
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  2. #2
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    #1.

    Do L'hopital twice.

    That is, take the derivative of the numerator and denominators twice

    \frac{d}{dx}[e^{x}-x-1]=e^{x}-1

    \frac{d}{dx}[e^{x}-1]=e^{x}

    The second derivative of x^{2} is 2.

    You have: \lim_{x\to\0}\frac{e^{x}}{2}

    Now, as x approaches 0, you have 1/2

    For #2. I believe you may have a 'tan' in your future.

    For #3. \frac{A}{y}+\frac{B}{y+2}+\frac{C}{y-3}=4y^{2}-7y-12

    A(y+2)(y-3)+B(y-3)y+C(y+2)y=4y^{2}-7y-12

    Ay^{2}+By^{2}+Cy^{2}-Ay-3By+2Cy-6A=4y^{2}-7y-12

    Equate coefficients:

    \left[\begin{array}{ccc|c}A&B&C&4\\-A&-3B&2C&-7\\-6A&0&0&-12\end{array}\right]<br />

    Solve the system.

    Soroban has a slick way of doing partial fractions. I trust he'll be along to demonstrate.
    Last edited by galactus; July 26th 2006 at 04:04 AM.
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  3. #3
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    Hello, c_323_h!

    Cody (Galactus) is right . . . there is a slick way.
    You may already know it, though.


    3. Partial Fractions: . \int \frac{4y^2-7y-12}{y(y+2)(y-3)}dy

    We have the set up: . \frac{4y^2 - 7y - 12}{y(y+2)(y-3)}\:=\:\frac{A}{y} + \frac{B}{y+2} + \frac{C}{y-3}

    Multiply through by the LCD:
    . . 4y^2 - 7y - 12 \:= \:A(y+2)(y-3) + By(y-3) + C(y(y-2)


    Now pick some "good" values for y *

    Let y = 0:\;\;-12\:=\:A(2)(-3) + B(0) + C(0)\quad\Rightarrow\quad-6A =  -12\quad\Rightarrow\quad \boxed{A = 2}

    Let y = -2:\;\;18 \:=\:A(0) + B(-2)(-5) + C(0)\quad\Rightarrow\quad10B =  18\quad\Rightarrow\quad \boxed{B = \frac{9}{5}}
    Let y = 3:\;\;3 \:=\:A(0) + B(0) + C(3)(5)\quad\Rightarrow\quad 15C =  3\quad\Rightarrow\quad \boxed{C = \frac{1}{5}}

    Therefore: . \frac{4y^2 - 7y - 12}{y(y+2)(y-3)} \:=\:\frac{2}{y} + \frac{\frac{9}{5}}{y + 2} + \frac{\frac{1}{5}}{y-3}

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    * I hope it's obvious that "good" values are
    . . those that create zeros in the expression.

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  4. #4
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    Quote Originally Posted by Soroban
    Hello, c_323_h!

    Cody (Galactus) is right . . . there is a slick way.
    You may already know it, though.



    We have the set up: . \frac{4y^2 - 7y - 12}{y(y+2)(y-3)}\:=\:\frac{A}{y} + \frac{B}{y+2} + \frac{C}{y-3}

    Multiply through by the LCD:
    . . 4y^2 - 7y - 12 \:= \:A(y+2)(y-3) + By(y-3) + C(y(y-2)


    Now pick some "good" values for y *

    Let y = 0:\;\;-12\:=\:A(2)(-3) + B(0) + C(0)\quad\Rightarrow\quad-6A =  -12\quad\Rightarrow\quad \boxed{A = 2}

    Let y = -2:\;\;18 \:=\:A(0) + B(-2)(-5) + C(0)\quad\Rightarrow\quad10B =  18\quad\Rightarrow\quad \boxed{B = \frac{9}{5}}
    Let y = 3:\;\;3 \:=\:A(0) + B(0) + C(3)(5)\quad\Rightarrow\quad 15C =  3\quad\Rightarrow\quad \boxed{C = \frac{1}{5}}

    Therefore: . \frac{4y^2 - 7y - 12}{y(y+2)(y-3)} \:=\:\frac{2}{y} + \frac{\frac{9}{5}}{y + 2} + \frac{\frac{1}{5}}{y-3}

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    * I hope it's obvious that "good" values are
    . . those that create zeros in the expression.

    yeah...this is the way they taught me in class. just couldn't quite the decomposition right. thanks

    thanks to galactus too
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  5. #5
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    Quote Originally Posted by c_323_h
    2. L'Hospital's Rule:
    \lim_{x\rightarrow{{\pi/2}^+}}\frac{cosx}{1-sinx}
    They give, \frac{0}{0} use L'HopitAl
    Derivative of numerator,
    -\sin x
    Derivative of denominator,
    -\cos x
    Thus,
    \lim_{x\to\pi/2^+}\frac{-\sin x}{-\cos x}=\lim_{x\to\pi/2^+}\tan x=+\infty
    ---
    I recommend to avoid L'Hopital but rather do,
    \frac{\cos x}{1-\sin x}\cdot \frac{1+\sin x}{1+\sin x}=\frac{\cos x(1+\sin x)}{1-\sin^2x}= \frac{\cos x(1+\sin x)}{\cos ^2x}=\frac{1+\sin x}{\cos x}=\sec x+\tan x\to \infty as x\to \pi/2^+
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