Example 7. Find $\lim _{x \rightarrow+\infty} \frac{7 x^{2}+6 x-3}{9 x^{3}+8 x^{2}-2}$.
Solution. By dividing the numerator and the denominator of the fraction by $x^{3}$, i.e., by the highest degree, we get

$$
\begin{aligned}
\lim _{x \rightarrow+\infty} \frac{7 x^{2}+6 x-3}{9 x^{3}+8 x^{2}-2}= & \lim _{x \rightarrow+\infty} \frac{\frac{7}{x}+\frac{6}{x^{2}}-\frac{3}{x^{3}}}{9+\frac{8}{x}-\frac{2}{x^{3}}}=\frac{\lim _{x \rightarrow+\infty}\left(\frac{7}{x}+\frac{6}{x^{2}}-\frac{3}{x^{3}}\right)}{\lim _{x \rightarrow+\infty}\left(9+\frac{8}{x}-\frac{2}{x^{3}}\right)}= \\
& =\frac{0+0-0}{9+0-0}=\frac{0}{9}=0
\end{aligned}
$$

Remark. To resolve the indeterminacy of the form $\frac{\infty}{\infty}$, given by the ratio of two polynomials, one must divide the numerator and the denominator by the highest power of $x$ present in them, and then take the limit.