unsure how to rearrange $f(x)$ into suitable $p(x)/q(x)$

Question

Consider the function $f(x)= (x^3 + 2x - 3) / (x^2 + 3x + 4)$

by dividing the numerator and denominator by the highest power of $x$ present, convert $f(x)$ into the form $P(x)/Q(x)$ where both $P(x)$ and $Q(x)$ have finite limits as '$x$ tends to infinity', not both $0$.

I know I am supposed to divide the numerator and denominator separately, however when it says 'by the highest power of $x$ present', does this mean

present in itself (ie numerator OR denominator) or the fuction $f(x)$ as a whole, (ie in this question $x^3$) I have assumed its the highest power of $x$ present in the whole function $f(x)$, and I have ended up with

$$ (1 + 2x^{-2} - 3x^{-3}) / (x^{-1} + 3x^{-2} + 4x^{-3}) $$

and I do not know where to go from here.

not entirely sure how to rearrange negative powers. ie, if theres a negative power

can I just swap that term from the top to the bottom or vice versa?

Answer 1

Divide not by $x^3$ but $x^2$ to get

$$\frac{x^3+2x-3}{x^2+3x+4}\underset{x>0}{=}\frac{x+2/x-3/x^2}{1+3/x+4/x^2}\underset{x\rightarrow\infty}{\sim} \frac{x}{1}=x.$$

$x$ is said to be an asymptotic of $f(x)$ and we write $f(x)\sim x$ (for $x$ large).

Answer 2

Your answer fits the literal instructions of the question. However, I'm not sure why it's a useful thing to do. Your question as stated does not ask you to evaluate the limit, so you're done.

If I were to write the question, I would say to do the following:

$$\frac{x^3 + 2x -1}{x^2 + 3x + 4} = \frac{x^3(1 + 2x^{-2} -x^{-3})}{x^2(1 + 3x^{-1} + 4x^{-2})} \sim x$$