Introductory Calculus: Limit as x goes to infinity

[jumbogala]

Homework Statement

What is the limit as x goes to positive infinity of:

(4x²+3x+8)/(6x²+5x-7)?

Homework Equations

The Attempt at a Solution

Usually I would factor the top and bottom and see if something cancels, but that doesn't work here.

Also l'hopital rule doesn't work.

I thought the answer would be one because substituting x for infinity gives infinity/infinity=1. But apparently that's not the correct answer.

Help?

 

[Dick]

l'Hopital's rule does work. But you don't need it. Just divide numerator and denominator by x^2.

 

[NBAJam100]

When you have something like (4x2+3x+8)/(6x2+5x-7) look to see if the numerator and denominators have matching highest x values (e.x- both top and bottom have x^2, which is in fact the value of x raised to the highest power in the equation). Then the limit is the fraction of the 2 leading numbers.

 

[mutton]

Infinity/infinity is an indeterminate form. Consider for example \lim_{x \to \infty} \frac{x}{x^2}. Both top and bottom approach infinity, but the limit is not 1.

 

[Mark44]

Multiply the whole rational expression by 1 in the form of
\frac{\frac{1}{x^2}}{\frac{1}{x^2}}
The limit should be pretty obvious after that.

 

[jumbogala]

Would you always divide the numerator and the denominator by the highest power?

Ex. If you had (x^5 + x^2) / (x^3 + x), would you divide by x^5?

Okay, so if I divide both numerator and denominator by x^2, I end up with a limit of 4/6.

 

[Mark44]

Or, reduced, 2/3.
On your question, dividing numerator and denominator by the highest power in either should work. That way, you'll get a constant + terms that go to zero in one place, and a variable to the resulting power in the other. The resulting expression will have a limit of infinity if the variable is in the numerator, and will be zero if the variable is in the denominator.

 

[jumbogala]

Let me make sure I understood that correctly. So for example,

limit as x goes to + infinity of (9x²-3x+4)/(2x-5)

- I divide everything by x squared
- After removing terms that go to zero I have (9x - 3)/(2)

Since the variable is in the numerator, the limit is infinity?

 

[Mark44]

You actually divided by x, which was the right thing to do (and not what I said earlier).

In the problem you just showed, dividing numerator and denominator by x gives:
\frac{9x -3 + 4/x}{2 - 5/x}

Now, if you take the limit as x approaches infinity, the numerator grows large without bound (the constant -3 can be ignored and the 4/x term goes to zero), and the denominator approaches 2 (the -5/x term goes to zero). So the whole rational expression approaches infinity.

Another approach when the degree of the polynomial in the numerator is greater than that of the polynomial in the denominator (as in your last example) is to divide the numerator by the denominator by long division. In this case you'll get 9/2*x + 39/4 + 211/[4(2x - 5)]. A little messy, but it's easy to see that as x gets large the overall expression also gets large, which is consistent with what you found earlier.

To set you on the right path in dealing with quotients of polynomials P(x)/Q(x):
If deg(P) = deg(Q), divide P(x) and Q(x) by the highest power of x in either.
If deg(P) < deg(Q), divide P(x) and Q(x) by the highest power of x in P.
If deg(P) > deg(Q), divide P(x) and Q(x) by the highest power of x in Q.

 

[NBAJam100]

Infinity/infinity is an indeterminate form. Consider for example \lim_{x \to \infty} \frac{x}{x^2}. Both top and bottom approach infinity, but the limit is not 1.

Yeah, that's not what i was saying at all though. Because X^1 and X^2 are different, if the highest leading x terms match though (such as X^2/X^2) then what i said would apply.

 

[NBAJam100]

Would you always divide the numerator and the denominator by the highest power?

Ex. If you had (x^5 + x^2) / (x^3 + x), would you divide by x^5?

Okay, so if I divide both numerator and denominator by x^2, I end up with a limit of 4/6.

You misunderstood, you don't divide by the leading power, you divide by the number in front of the leading powers (and it only applies if the x raised to the highest power is present in both the numerator and the denominator, then you apply by the leading numbers...)

Ex Lim x--> infinity of (x^7+x^5)/(7x^7+3) ----> 1/7, because x^7 was the highest raised power and was present in both numerator and denominator and 1/7 are the leading numbers in front of the leading x^n... its just a short cut to the actual way of doing it, sorry if i confused you

 

[Mark44]

You misunderstood, you don't divide by the leading power, you divide by the number in front of the leading powers (and it only applies if the x raised to the highest power is present in both the numerator and the denominator, then you apply by the leading numbers...)

No, that isn't right. In the case you're describing, in which the degrees of the numerator and denominator polynomials are the same, you divide both polynomials by the highest power of the variable in either place, not by the coefficients. When you do this you'll be left with are constants plus terms that go to zero in the numerator and denominator.

Ex Lim x--> infinity of (x^7+x^5)/(7x^7+3) ----> 1/7, because x^7 was the highest raised power and was present in both numerator and denominator and 1/7 are the leading numbers in front of the leading x^n... its just a short cut to the actual way of doing it, sorry if i confused you

 

[NBAJam100]

No, that isn't right. In the case you're describing, in which the degrees of the numerator and denominator polynomials are the same, you divide both polynomials by the highest power of the variable in either place, not by the coefficients. When you do this you'll be left with are constants plus terms that go to zero in the numerator and denominator.

Uhhh... I am not saying you divide the whole equation by the coefficients, I am saying the answer is the ratio of leading coefficients of the numerator/denominator.

 

[Mark44]

Uhhh... I am not saying you divide the whole equation by the coefficients, I am saying the answer is the ratio of leading coefficients of the numerator/denominator.

NBAJam100,
Here's what you said a couple of posts back:

you divide by the number in front of the leading powers

The number in front of the leading power is the coefficient

 

[NBAJam100]

NBAJam100,
Here's what you said a couple of posts back:

The number in front of the leading power is the coefficient

I admit, i worded it a little awkwardly (very awkwardly... ) I didnt mean divide the whole equation by it, i meant divide the num and denom numbers... either way, sorry for the confusion.