When does infinity apply? Find asymptotes of 4x^2 + 5xy + y^2 = 10

kinosh

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Jan 16, 2018
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Problem: Find the asymptotes of the hyperbola 4x^2 + 5xy + y^2 = 10.

The book says to rearrange this equation to y^2 + (5x)y + (4x^2 - 10) = 0, then use the quadratic formula, (-5x +/- sqrt(25x^2 - 4 (4x^2 - 10)))/2. Simplifying a bit to get
(-5x +/- sqrt (9x^2 - 40))/2. Then the book factors out 3x from the sqrt to get (-5x +/- 3x sqrt(1 - 40/9x^2))/2. I got it up to this point.

Then the book says the 40/9x^2 term approaches zero as x -> inf, which I agree with, to get y = (-5x +/- 3x)/2. Why are the -5x and 3x terms ignored when applying the x -> inf?
 

j-astron

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Jan 10, 2018
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Problem: Find the asymptotes of the hyperbola 4x^2 + 5xy + y^2 = 10.

The book says to rearrange this equation to y^2 + (5x)y + (4x^2 - 10) = 0, then use the quadratic formula, (-5x +/- sqrt(25x^2 - 4 (4x^2 - 10)))/2. Simplifying a bit to get
(-5x +/- sqrt (9x^2 - 40))/2. Then the book factors out 3x from the sqrt to get (-5x +/- 3x sqrt(1 - 40/9x^2))/2. I got it up to this point.

Then the book says the 40/9x^2 term approaches zero as x -> inf, which I agree with, to get y = (-5x +/- 3x)/2. Why are the -5x and 3x terms ignored when applying the x -> inf?
Your book is saying that the \(\displaystyle 40/9x^2 \) term actually approaches zero as x gets larger and larger. So, its contribution to the expression can be ignored for really large values of x, because it will be so close to zero as to be negligible. The other terms, however, do not approach zero as x gets larger and larger. Rather, they also get very very large. So they are not ignored. They are retained. Does that make sense?
 

JeffM

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Sep 14, 2012
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Problem: Find the asymptotes of the hyperbola 4x^2 + 5xy + y^2 = 10.

The book says to rearrange this equation to y^2 + (5x)y + (4x^2 - 10) = 0, then use the quadratic formula, (-5x +/- sqrt(25x^2 - 4 (4x^2 - 10)))/2. Simplifying a bit to get
(-5x +/- sqrt (9x^2 - 40))/2. Then the book factors out 3x from the sqrt to get (-5x +/- 3x sqrt(1 - 40/9x^2))/2. I got it up to this point.

Then the book says the 40/9x^2 term approaches zero as x -> inf, which I agree with, to get y = (-5x +/- 3x)/2. Why are the -5x and 3x terms ignored when applying the x -> inf?
They are NOT ignored.

What is being said is that

\(\displaystyle y \approx \dfrac{-\ 5x + 3x}{2} = -\ x \text { or } y \approx {-\ 5x - 3x}{2} = -\ 4x.\)

\(\displaystyle |x| >> 0 \implies x \text { is VERY SIGNIFICANT.}.\)
 

kinosh

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I'm still confused. I understand that 1/x -> 0 as x -> inf. But why doesn't a term like 5x -> inf at the same time, especially if 1/x and 5x are in the same equation? It's like saying 1/x + 5x -> 5x for x -> inf instead of 1/x + 5x being undefined.
 

j-astron

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I'm still confused. I understand that 1/x -> 0 as x -> inf. But why doesn't a term like 5x -> inf at the same time, especially if 1/x and 5x are in the same equation? It's like saying 1/x + 5x -> 5x for x -> inf instead of 1/x + 5x being undefined.
Why would it be undefined? The two terms are being added. That doesn't lead to 0/0 or something else undefined.

(Really really small number) + (really really large number) is basically just going to be equal to the really really large number.

Suppose x = 1 million

5x + 1/x = 5,000,000 + 0.000001

You can pretty safely ignore the contribution from the second term here already. EDIT: and if you make x even larger, then you can even more safely ignore the second term.
 
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JeffM

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Sep 14, 2012
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I'm still confused. I understand that 1/x -> 0 as x -> inf. But why doesn't a term like 5x -> inf at the same time, especially if 1/x and 5x are in the same equation? It's like saying 1/x + 5x -> 5x for x -> inf instead of 1/x + 5x being undefined.
If you think about what "approach" means in mathematics, it is a little deceptive. When we say "x approaches a," where a is a real number, we mean that x APPROXIMATES BUT DOES NOT EQUAL a. When, however, we say "x approaches infinity" or "x approaches negative infinity," we can't mean the same thing because no finite number approximates infinity. Instead we mean "x is a finite number with a very large absolute value."

OK. So when we say

\(\displaystyle \text {As x approaches infinity, } \dfrac{1}{x} + 5x \text { approaches } 5x\),

what we mean is almost exactly what you said:

\(\displaystyle \text {As x gets very large, } \dfrac{1}{x} + 5x \text { ALMOST equals } 5x.\)

Standard analysis, a mathematical theory underlying calculus, makes this vocabulary precise, but if you realize that "approaching a finite number" means "approximating that finite number" but "approaching infinity" means "picking a finite number with a large absolute value," limits will make intuitive sense.
 
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