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Finding the limit of a quotient

Given the function

f(x)=2xx2+x−2f(x)=\dfrac{2x}{x^2+x-2}

I need to find the horizontal asymptote, so dividing each term by the highest power (x^2) in the denominator gives

f(x)=2x1+1x−2x2f(x)=\dfrac{\frac{2}{x}}{1+\frac{1}{x}-\frac{2}{x^2}}

so

lim⁡x→∞f(x)=0\displaystyle \lim_{x \to \infty} f(x) = 0

But if I factorise the denominator first,

f(x)=2x(x+2)(x−1)f(x)=\dfrac{2x}{(x+2)(x-1)}

and divide each term (this time the highest power is x)

f(x)=2(1+2x)(1−1x)f(x)=\dfrac{2}{(1+\frac{2}{x})(1-\frac{1}{x})}

you get a different limit

lim⁡x→∞f(x)=2\displaystyle \lim_{x \to \infty} f(x) = 2

So why is the second method wrong?
(edited 12 years ago)
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Avatar for Hopple
Hopple
18
In the second method, you've essentially divided the denominator by x^2, whereas only divided the numerator by x.
Original post by Plato's Trousers
Given the function

f(x)=2xx2+x−2f(x)=\dfrac{2x}{x^2+x-2}

I need to find the horizontal asymptote, so dividing each term by the highest power (x^2) in the denominator gives

f(x)=2x1+1x−2x2f(x)=\dfrac{\frac{2}{x}}{1+\frac{1}{x}-\frac{2}{x^2}}

so

lim⁡x→∞f(x)=0\displaystyle \lim_{x \to \infty} f(x) = 0

But if I factorise the denominator first,

f(x)=2x(x+2)(x−1)f(x)=\dfrac{2x}{(x+2)(x-1)}

and divide each term (this time the highest power is x)

f(x)=2(1+2x)(1−1x)f(x)=\dfrac{2}{(1+\frac{2}{x})(1-\frac{1}{x})}

you get a different limit

lim⁡x→∞f(x)=2\displaystyle \lim_{x \to \infty} f(x) = 2

So why is the second method wrong?

1x(x+2)(x−1)≡(1+2x)(x−1)≡(x+2)(1−x2)≢(1+2x)(1−x2)\dfrac{1}{x}(x+2)(x-1) \equiv (1+\frac{2}{x})(x-1)\equiv (x+2)(1-\frac{x}{2}) \not\equiv (1+\frac{2}{x})(1-\frac{x}{2})
Original post by Hopple
In the second method, you've essentially divided the denominator by x^2, whereas only divided the numerator by x.




Original post by Farhan.Hanif93
1x(x+2)(x−1)≡(1+2x)(x−1)≡(x+2)(1−x2)≢(1+2x)(1−x2)\dfrac{1}{x}(x+2)(x-1) \equiv (1+\frac{2}{x})(x-1)\equiv (x+2)(1-\frac{x}{2}) \not\equiv (1+\frac{2}{x})(1-\frac{x}{2})


Ah... yes. That'll be it then. :stupido:

:getmecoat:
(edited 12 years ago)
Original post by Plato's Trousers
...


You probably don't care anymore but it's worth noting that this is a quadratic in x so you can find the horizontal asymptotes by considering the discriminant.
Original post by ben-smith
You probably don't care anymore but it's worth noting that this is a quadratic in x so you can find the horizontal asymptotes by considering the discriminant.


No, I do still care. Could you explain?
Original post by Plato's Trousers
No, I do still care. Could you explain?


y=2xx2+x−2⇒2x=yx2+yx−2y[br]⇒yx2+(y−2)x−2y=0y=\dfrac{2x}{x^2+x-2} \Rightarrow 2x=yx^2+yx-2y[br]\Rightarrow yx^2+(y-2)x-2y=0
Now consider the nature of the roots of this polynomial in terms of y.
Original post by ben-smith
y=2xx2+x−2⇒2x=yx2+yx−2y[br]⇒yx2+(y−2)x−2y=0y=\dfrac{2x}{x^2+x-2} \Rightarrow 2x=yx^2+yx-2y[br]\Rightarrow yx^2+(y-2)x-2y=0
Now consider the nature of the roots of this polynomial in terms of y.


So in the determinant,

b2=(y−2)2=y2−4y+4b^2 =(y-2)^2=y^2-4y+4

and

4ac=−8y24ac=-8y^2

So the roots are real (since b2>4acb^2 > 4ac)

not sure what that has to do with the horizontal asymptote though

EDIT: hang on.. is this because putting the equation in terms of y basically switches the axes, so we are basically finding the vertical asymptotes?
(edited 12 years ago)

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