Help, difficult math problem! How did you do this problem and what rules did you use?

Question

Let f(x) = a sub n x^n + a sub n-1 x^n-1 +...+ a sub 2 x^2 + a sub 1 x + a sub 0

and

g(x) = b sub k x^k + b sub k-1 x ^k-1 + ... + b sub 2 x^2 + b sub 1 x + b sub 0 .

sub=subscript

n and k are whole numbers. All of the coefficients are real and the lead coefficients (the coefficients of the highest powers) are positive.

Then if n > k, lim f(x)/g(x) =
.... x=>oo

I have to chose from

... a sub n / b sub k
... a sub 0 / b sub 0
... oo
... 0

or is it none of these

Answer 1

When X=>oo lim f(x)/g(x) = a sub 0 / b sub 0

This appears to be a rational polynomial infinity limit problem.

You can go at this page to view the rule:
http://72.14.207.104/search?q=cache:JTZNye14n84J:mathpost.asu.edu/~awtrey/MAT210fall06/Limits.pdf+rational+polynomial+limit&hl=en&gl=ca&ct=clnk&cd=2

Answer 2

The answer it quite simple really. With these types of problems, you can ignore everything but the two nomials with the highest coefficients, the one on top and the one on bottom. So you can rewrite it as

lim as x appoaches 0 of (a sub n times x^n)/(b sub k times x^k)

you said n was greater than k , meaning that the numerator is larger than the denominator.

.....I think thats the kind of work your supposed to show but in the end i think it should just be neither, because it seems to me it is undefined.

I hope this helps, its been awhile since ive done this stuff

Answer 3

If n>k, a sub n >0 and x=>oo, lim f(x)/g(x) will be oo.

If x becomes very large the answer will be determined by the highest power in the division. Here: a sub n x^n.
Since lim x=>oo a sub n x^n is oo, lim f(x)/g(x) will also be oo.

Answer 4

the rule is that for any eps > 0 there exists an N >0 such that for all x >N |g(x)/f(x)| < eps

and thus f(x)/g(x) -> oo as x ->00 , (since a_n,b_n >0)

Answer 5

a_0 / b_0 as long as b_0 isn't 0

The limit of the quotient is the quotient of the limits.
The two functions are polynomials so lim x->0 f(x) = f(0) = a_0
and lim x->0 g(x) = g(0) = b_0