ACK!!! I have a math question. . .

[Scrapster]

An example out of our book tells us how to find the limit of the equation:

2x^3 + x^2 - 7x
---------------
x^3 + 2x + 2

Now, they did some kind of trick and transformed the above equation into the following:

2 + (1/x) - (7/x^2)
--------------------
1 + (2/x^2) + (2/x^3)

And then they just had x go to infinity and you are left with a limit of 2. But I don't know how they did the trick in the middle to transform the equation into a better looking equation. Does someone here know what they did?

 

[hendon]

they just divided numerator and denominator by the highest power of x

 

[gwlam12]

agreed. everything is divided by x^3

 

[Scrapster]

say again?

 

[GL]

Just to generalize the approach, simply divide the numerator and denomator by the highest degree 'x' term.

oops...didn't see Hendon's post.

The reason they just got 2 as the limit is because when you're dealing with high degree 'x' terms, the lower degree 'x' terms become irrelevant. For example, as you approach infinity, x^3 + x and x^3 + 2x are more affected by the x^3 term than the x terms.

 

[Scrapster]

x^3? Why is that? Why isn't it divided by 2x^3?

 

[hendon]

<< x^3? Why is that? Why isn't it divided by 2x^3? >>



If you do that, you get the same answer

 

[MrCodeDude]

No, becuase if you divide by 2X^3 you get:

1 + (1/2x) - 3.5x^2
-------------------
2 + (1/x^2) + x^3

Isn't 2/1 = 2; but 1/2 = .5?
Or you can't divide until you simplify the equation?
-- mrcodedude

 

[hendon]

No, it's the same
because if you divide by 2x^3
it becomes:
1+ 1/2x - 7/2x^2
---------------
1/2 + 1/x + 2/x^2

which is the same

 

[goodoptics]

agreed with hendon.

 

[Mday]

i also agree with whatever his name is.

--

there are several ways of doing this problem depending on what strategies you want to use.

l'hopital's rule is one option (you have to take the derivative several times, which is easy enough.

or you can separate the fraction into 3 fractions (which have the same denominator), 2 of those fractions approach 0 as x -> inf.