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Math Help - isoclines,,,never been very good with limits

  1. #1
    Member i_zz_y_ill's Avatar
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    isoclines,,,never been very good with limits

    can u cancel infinty,,,,this is probably a very stupid/petty question!!Q is find the equation of the straight line on which the gradient of the solution curve is infinite. I made the presumption that this straight line would essentially be an isocline(hopefulli im ryt) and did the follwing,, Is is more accurate using the D.e and making bottom=0???Or is my working ok?!

    dy/dx=(3y-x)/(y+x)

    dy/dx=m(constant)

    putting constant as infinity and rearranging to f(x)

    (i) y= -x(1-m)/(m-3)
    (ii) y= -x(1-inf)/(inf-3) implies(i think) y= -x(inf)/(inf)
    y = -x

    is this okai???
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by i_zz_y_ill View Post
    can u cancel infinty,,,,this is probably a very stupid/petty question!!Q is find the equation of the straight line on which the gradient of the solution curve is infinite. I made the presumption that this straight line would essentially be an isocline(hopefulli im ryt) and did the follwing,, Is is more accurate using the D.e and making bottom=0???Or is my working ok?!

    dy/dx=(3y-x)/(y+x)

    dy/dx=m(constant)

    putting constant as infinity and rearranging to f(x)

    (i) y= -x(1-m)/(m-3)
    (ii) y= -x(1-inf)/(inf-3) implies(i think) y= -x(inf)/(inf)
    y = -x

    is this okai???
    If your question is

    \lim_{m\to\infty}x\frac{1-m}{m-3}

    Then we would have as you observed

    \frac{1-m}{m-3}\sim\frac{-m}{m}=-1

    giving us

    \lim_{m\to\infty}x\frac{1-m}{m-3}=-x

    EDIT: Please, for your sake never make this connection

    \lim_{x\to\infty}\frac{f(x)}{g(x)}=\frac{\infty}{\  infty}\Rightarrow{\lim_{x\to\infty}\frac{f(x)}{g(x  )}=1}


    Counter example

    \lim_{x\to\infty}\frac{x}{e^{10^{10^{10^x}}}}=\fra  c{\infty}{\infty}

    But

    \lim_{x\to\infty}\frac{x}{e^{10^{10^{10^x}}}}=0
    Last edited by Mathstud28; June 7th 2008 at 07:44 PM.
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  3. #3
    Member i_zz_y_ill's Avatar
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    nah i think the m value should just equal infinty. If you imagine all the direction indicators on a tangent field being vertical in which case i believe there'll be no limits excuse my wording earlier. Erm my query was more in this situation by putting m = inf does this diminish the accuracy of my working when i say that y= -x(1+inf)/(inf - 3) is the equivalent of y= -x(inf/inf) in which case inf would cancel leaving me witht the right answer x + y = 0




    oh ryt:P errrr well isnt it f(x)/f(x) where the bottom is only a constnt so im not reli doing that. Iim still confsed as to whether i should use limits or not?? lol i should just equate the bottom to zero
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by i_zz_y_ill View Post
    nah i think the m value should just equal infinty. If you imagine all the direction indicators on a tangent field being vertical in which case i believe there'll be no limits excuse my wording earlier. Erm my query was more in this situation by putting m = inf does this diminish the accuracy of my working when i say that y= -x(1+inf)/(inf - 3) is the equivalent of y= -x(inf/inf) in which case inf would cancel leaving me witht the right answer x + y = 0
    It is very hard to understand what you are saying. But if you are asking can you just say that since m tends towards infinity that the quotient of the m terms is -1, yes. But if you will look above cancelling infinities is a mathematically unforgivable act
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  5. #5
    Member i_zz_y_ill's Avatar
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    k cheers i think i can see where im wrong now(didtn read q), i'll ook out for that error aswell !!!thnx
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