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Math Help - Calc: L'Hopital's Rule

  1. #1
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    Calc: L'Hopital's Rule

    hey can someone please see if I did this right,

    x--> infinity, (4x+9) / (7x^2 +4x - 9)
    take the derivative = 4 / 14x + 4
    so would it be 0? if not i'm not sure what to do next.


    and

    the integral from 0 to infinity of 1 / (x(x^2)+5)
    so i did 1 / x^3 + 5x then i can't take the derivative again b/c the numerator is 1. now i don't know what to do.

    any help is much appreciated.
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  2. #2
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    Quote Originally Posted by darkblue View Post
    hey can someone please see if I did this right,

    x--> infinity, (4x+9) / (7x^2 +4x - 9)
    take the derivative = 4 / 14x + 4
    so would it be 0? if not i'm not sure what to do next..
    No need to use it.
    Use divide by x^2 the numerator and denominator.
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by darkblue View Post
    hey can someone please see if I did this right,

    x--> infinity, (4x+9) / (7x^2 +4x - 9)
    take the derivative = 4 / 14x + 4
    so would it be 0? if not i'm not sure what to do next.
    yes, it would be zero.



    the integral from 0 to infinity of 1 / (x(x^2)+5)
    so i did 1 / x^3 + 5x then i can't take the derivative again b/c the numerator is 1. now i don't know what to do.

    any help is much appreciated.
    ok. you have two different things here. you said you had 1/(x(x^2) + 5), but if you expand that, you would have 1/(x^3 + 5) not 1/(x^3 + 5x). so i think you made a typo here. please correct it
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  4. #4
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    Quote Originally Posted by Jhevon View Post
    yes, it would be zero.


    ok. you have two different things here. you said you had 1/(x(x^2) + 5), but if you expand that, you would have 1/(x^3 + 5) not 1/(x^3 + 5x). so i think you made a typo here. please correct it
    thank you.
    oh i see. it is: 1 over x(x^2 + 5), then i just distributed the x.
    so 1 over (x^3 + 5x) ?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by darkblue View Post
    thank you.
    oh i see. it is: 1 over x(x^2 + 5), then i just distributed the x.
    so 1 over (x^3 + 5x) ?
    Don't distribute the x

    By the method of partial fractions:
    \frac {1}{x \left( x^2 + 5 \right)} = \frac {A}{x} + \frac {Bx + C}{x^2 + 5}

    \Rightarrow 1 = (A + B)x^2 + Cx + 5A

    \Rightarrow A + B = 0
    ........... C = 0
    .......... 5A = 1

    So we have: A = \frac {1}{5} \mbox { , } B = - \frac {1}{5} \mbox { and } C = 0

    So, \int \frac {1}{x \left( x^2 + 5 \right)}dx = \int \left( \frac { \frac {1}{5}}{x} - \frac { \frac {1}{5}x}{x^2 + 5} \right)dx

    = \frac {1}{5} \int \frac {1}{x} dx - \frac {1}{5} \int \frac {x}{x^2 + 5} dx

    Can you take it from here?
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  6. #6
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    ohh Partial Fractions! I forgot about that. thankyou

    so i got (1/5 ln x - (1/10) ln(x^2 + 5) + C
    is the 1/10 right? because i multiplied 1/5 by 1/2.
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by darkblue View Post
    ohh Partial Fractions! I forgot about that. thankyou

    so i got (1/5 ln x - (1/10) ln(x^2 + 5) + C
    is the 1/10 right? because i multiplied 1/5 by 1/2.
    yes, that's correct
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  8. #8
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    thanx again! problem solved!
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