# isoclines,,,never been very good with limits

• Jun 7th 2008, 06:25 PM
i_zz_y_ill
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???
• Jun 7th 2008, 06:32 PM
Mathstud28
Quote:

Originally Posted by i_zz_y_ill
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???

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

Then we would have as you observed

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

giving us

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

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

Counter example

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

But

$\displaystyle \lim_{x\to\infty}\frac{x}{e^{10^{10^{10^x}}}}=0$
• Jun 7th 2008, 06:43 PM
i_zz_y_ill
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
• Jun 7th 2008, 06:47 PM
Mathstud28
Quote:

Originally Posted by i_zz_y_ill
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
• Jun 7th 2008, 06:53 PM
i_zz_y_ill
k cheers i think i can see where im wrong now(didtn read q:D), i'll ook out for that error aswell !!!thnx