Completing Limit Equation

[reardear]

Hi, I've been trying to figure this one out for a while now. I know that a positive or negative #/0 = infinity, so that's what I've been trying to do with the first coefficients. I guess I'm probably not approaching this the right way. Any tips would be greatly appreciated.

lim
x->∞

9x⁸-15x²+6 / cxⁿ-8x²+15 = ∞
where c =/= 0

I have to find what:
- number C is bigger or smaller than
- number N is bigger, equal to, or smaller than.

 

[JeffM]

Hi, I've been trying to figure this one out for a while now. I know that a positive or negative #/0 = infinity, so that's what I've been trying to do with the first coefficients. I guess I'm probably not approaching this the right way. Any tips would be greatly appreciated.

lim
x->∞

9x⁸-15x²+6 / cxⁿ-8x²+15 = ∞
where c =/= 0

I have to find what:
- number C is bigger or smaller than
- number N is bigger, equal to, or smaller than.

Is this the problem; under what conditions does

\(\displaystyle \displaystyle \lim_{x \rightarrow \infty}\dfrac{9x^8 - 15x^2 + 6}{cx^n - 8x^2 + 15} = \infty\ if\ c \ne 0.\)

I think this is easiest to think about intuitively.

If c < 0 and x gets very large in the positive direction, the numerator becomes positive and the denominator becomes negative so the limit cannot possibly be positive infinity. So c > 0.

What happens when n > 8?

When n < 8?

If n = 8, you may need to think more deeply about c.

 

[soroban]

domain

Hello, reardear!

We can "eyeball" the problem.
Is that permitted?

\(\displaystyle \displaystyle\lim_{x\to\infty}\frac{9x^8 - 15x^2 + 6}{cx^n - 8x^2 + 15} \;=\;\infty\;\;\text{ where }c \ne 0.\)

\(\displaystyle \text{I have to find: }\:\begin{Bmatrix} \text{the domain of }n. \\ \text{the domain of }c. \end{Bmatrix}\)

Since the limit goes to \(\displaystyle \infty,\:n < 8.\)

Since the limit goes to \(\displaystyle +\infty,\:c > 0.\)

 

[HallsofIvy]

As x goes to infinity, the "leading term" of a polynomial is what is important.

That is, to find the limit, as x goes to infinity, of \(\displaystyle \frac{9x^8- 15x^2+ 6}{cx^n- 8x^2+ 15}\), we look at \(\displaystyle \frac{9x^8}{cx^n}\). If n> 8, that is \(\displaystyle \frac{9}{c}\frac{1}{x^{n- 8}}\). If n< 8, that is \(\displaystyle \frac{9}{c}x^{8- n}\). And, of course, if n= 8, \(\displaystyle \frac{9}{c}\).