Proving the Limit of f(x)-g(x) for Even Integer Polynomials

[madah12]

Homework Statement

I want to prove that a polynomial f(x) and a polynomial g(x) with degrees of k,n where k,n are positive even integer, n>k
that limit x-> - infinity of f(x)-g(x)=-infinity

Homework Equations

a polynomial can be written as
a₁x^n+a₂x^(n-1)...+a_(n-1)x+a_n

The Attempt at a Solution

Since n>k and k,n are positive even integers there is a positive even integer C such that
n=(k+C)
therefore we write f(x)=a₁x^k+a₂x^(k-1)+...+a_(k-1)x+a_k/b₁x^(c+k)
g(x)=b₁x^(k+C)+b₂x^(k+C)-1+...+b_(k+c-1)x+b_(k+c)
f(x)-g(x)=[a₁x^k+a₂x^(k-1)+...+a_(k-1)x+a_k]-[b₁x^(k+C)+b₂x^((k+C)-1)+...+b_(k+c-1)x+b_(k+c)]
=b₁x^(k+c)[[a₁/(b₁*x^c) + a₂/(b1*x^(-c-1)) ...a_(k-1)/b₁x^(c+k-1) + a_k/b₁x^(c+k)] -[ 1+b₂/b₁x +...+b_(c+k-1)/b₁x^(k+c-1)+b_c+k/b₁x^(c+k)]]
as x->-infinity
c/x^n = 0 THEOREM
therefore
since k+c is even then lim x->-infinity x^(c+K)=infinity
so limit x->-infinity f(x)-g(x)
=infinity *[(0+0+...+0+0)-(1+0...+0+0)=infinity * -1 = -infinity.
I know this isn't any kind of rigorous proof so how can I improve it?

 

[madah12]

would this also qualify as proof for limit x->-infinity of g(x)-f(x) = infinity?

 

[madah12]

any help?

 

[Mark44]

It's not true, in general. The behavior of a polynomial depends on both its degree and the sign of the highest degree coefficient.

For example,
[tex]\lim_{x \to -\infty} 3x^2 = +\infty[/tex]

but
[tex]\lim_{x \to -\infty} -2x^2 = -\infty[/tex]

For the theorem you are trying to prove, g(x) is of higher degree than f(x), so g(x) is going to dominate f(x) in the difference f(x) - g(x), when x is very large or very negative.

Here are a couple of specific examples:
[tex]\lim_{x \to -\infty} x^2 - 2x^4= -\infty[/tex]

[tex]\lim_{x \to -\infty} x^2 - (-3x^4 + x^3)= +\infty[/tex]

 

[madah12]

but it is true if I assume both a1 and b1 to be positive?
Also I am assuming both polynomials have positive even degrees
so I am safe from the argument that 0 is a polynomial.

 

[Mark44]

Homework Statement

I want to prove that a polynomial f(x) and a polynomial g(x) with degrees of k,n where k,n are positive even integer, n>k
that limit x-> - infinity of f(x)-g(x)=-infinity

Homework Equations

a polynomial can be written as
a₁x^n+a₂x^(n-1)...+a_(n-1)x+a_n

The Attempt at a Solution

Since n>k and k,n are positive even integers there is a positive even integer C such that
n=(k+C)
therefore we write f(x)=a₁x^k+a₂x^(k-1)+...+a_(k-1)x+a_k/b₁x^(c+k)

What do you mean "therefore"? You should first define f(x) and g(x). A nicer way to do things is to have the indexes of the coefficients match the exponents on the variables. For example, you could define f(x) as
f(x) = a_kx^k + a_k-1x^k-1 + ... + a₁x + a₀, with a similar definition for g(x).

Why do you have this term at the end? a_k/b₁x^(c+k)

g(x)=b₁x^(k+C)+b₂x^(k+C)-1+...+b_(k+c-1)x+b_(k+c)
f(x)-g(x)=[a₁x^k+a₂x^(k-1)+...+a_(k-1)x+a_k]-[b₁x^(k+C)+b₂x^((k+C)-1)+...+b_(k+c-1)x+b_(k+c)]
=b₁x^(k+c)[[a₁/(b₁*x^c) + a₂/(b1*x^(-c-1)) ...a_(k-1)/b₁x^(c+k-1) + a_k/b₁x^(c+k)] -[ 1+b₂/b₁x +...+b_(c+k-1)/b₁x^(k+c-1)+b_c+k/b₁x^(c+k)]]
as x->-infinity
c/x^n = 0 THEOREM
therefore
since k+c is even then lim x->-infinity x^(c+K)=infinity
so limit x->-infinity f(x)-g(x)
=infinity *[(0+0+...+0+0)-(1+0...+0+0)=infinity * -1 = -infinity.
I know this isn't any kind of rigorous proof so how can I improve it?

 

[madah12]

uhm actually because when I was making the post I mistaken that line with the next I didn't notice this ak/b1x^(c+k)
Also about the therefore I meant that the general form of a polynomial is that of what I wrote.
Why would it matter if I started at a₀or started at a_c+k?

 

[Mark44]

Start by defining f(x) and g(x). You can index the coefficients either way, but it makes it easier to follow if the coefficient index matches the exponent.

 

[madah12]

ok then f(x) ,g(x) are polynomials
f(x)=a_kx^k+a_k-1x^(k-1)+...+a₁x+a₀
g(x)=b_(c+k)x^(c+k) +b_(c+k)-1x^(c+k-1)+...+b₁x+b₀

 

[Mark44]

Then f(x) - g(x) = a_kx^k+a_k-1x^(k-1)+...+a₁x+a₀ - (b_(c+k)x^(c+k) +b_(c+k)-1x^(c+k-1)+...+b₁x+b₀)

You can factor x^k out of all terms and then take your limit.

 

[madah12]

shouldnt I factor b_c+kx^(c+k)?
so I get b_c+kx^(c+k) [(0-0-...-00-0)-(1-0-0-0)
=infinity (-1)=-infinity?
because everything else will have an x in the denominator and as x-> infinity all will go to 0.

 

[Mark44]

OK, yes, you could do that.