There's a trick to this type of integral. As the limit approaches infinity, the leading terms end up being the only important ones because they go off to infinity the fastest. That is, for very large xx we have

\(\frac{x^{9}-4x^{5}+2x-13}{-3x^{9}+x^{8}-5x^{2}+2x}≈ \frac{7x^{9}}{-3x^{9}}\)

Because limits are all about what happens as these types of approximations become more and more accurate, it turns out that in the limit, the two expressions are exactly equal. There is likely a theorem in your textbook to this effect.

\(\lim_{x \rightarrow \infty} \frac{x^{9}-4x^{5}+2x-13}{-3x^{9}+x^{8}-5x^{2}+2x}\)

\(=\lim_{x \rightarrow \infty} \frac{7x^{9}}{-3x^{9}}\)

The benefit here is that that second limit is super easy to evaluate.

\(\lim_{x \rightarrow \infty} \frac{7x^{9}}{-3x^{9}} = -(7/3)\)

So you end up just getting a fraction of the leading coefficients.

Now here's a question for you to try to think through. What do you think would change if the degree of the numerator and denominator were not the same?

\(\frac{x^{9}-4x^{5}+2x-13}{-3x^{9}+x^{8}-5x^{2}+2x}≈ \frac{7x^{9}}{-3x^{9}}\)

Because limits are all about what happens as these types of approximations become more and more accurate, it turns out that in the limit, the two expressions are exactly equal. There is likely a theorem in your textbook to this effect.

\(\lim_{x \rightarrow \infty} \frac{x^{9}-4x^{5}+2x-13}{-3x^{9}+x^{8}-5x^{2}+2x}\)

\(=\lim_{x \rightarrow \infty} \frac{7x^{9}}{-3x^{9}}\)

The benefit here is that that second limit is super easy to evaluate.

\(\lim_{x \rightarrow \infty} \frac{7x^{9}}{-3x^{9}} = -(7/3)\)

So you end up just getting a fraction of the leading coefficients.

Now here's a question for you to try to think through. What do you think would change if the degree of the numerator and denominator were not the same?