Question

evaluate the limit \\(\lim_{x \to +\infty} \frac{9x^3 - 10x^2 - 2x}{9 - 5x - 9x^3}\\)

Explanation:

Step1: Divide numerator and denominator by \(x^3\)

For the numerator \(9x^3 - 10x^2 - 2x\), dividing each term by \(x^3\) gives \(\frac{9x^3}{x^3}-\frac{10x^2}{x^3}-\frac{2x}{x^3}=9 - \frac{10}{x}-\frac{2}{x^2}\).
For the denominator \(9 - 5x - 9x^3\), dividing each term by \(x^3\) gives \(\frac{9}{x^3}-\frac{5x}{x^3}-\frac{9x^3}{x^3}=\frac{9}{x^3}-\frac{5}{x^2}-9\).

Step2: Evaluate the limit as \(x\to\infty\)

As \(x\to\infty\), terms with \(x\) in the denominator (\(\frac{10}{x}\), \(\frac{2}{x^2}\), \(\frac{9}{x^3}\), \(\frac{5}{x^2}\)) approach \(0\).
So the numerator limit is \(9 - 0 - 0 = 9\), and the denominator limit is \(0 - 0 - 9=-9\).

Step3: Find the final limit

The limit becomes \(\frac{9}{-9}=-1\).

Answer:

\(-1\)