Question

Use the limit theorem and the properties of limits to find the limit.

Answer 1

Answer:

$b. \frac{2}{3}$

Step-by-step explanation:

The given limit is

$\lim_{x \to \infty} \frac{-2x^3+7x+2}{-3x^3-x+2}$

We divide the numerator and the denominator by $x^3$

$\lim_{x \to \infty} \frac{\frac{-2x^3}{x^3}+\frac{7x}{x^3}+\frac{2}{x^3}}{\frac{-3x^3}{x^3}-\frac{x}{x^3}+\frac{2}{x^3}}$

This simplifies to;

$\lim_{x \to \infty} \frac{-2+\frac{7}{x^2}+\frac{2}{x^3}}{-3-\frac{1}{x^3}+\frac{2}{x^2}}$

Apply the following limit property;

As $x\rightarrow -\infty, \frac{c}{x^n} \rightarrow 0$, where c is a constant.

This implies that;

$\lim_{x \to \infty} \frac{-2+\frac{7}{x^2}+\frac{2}{x^3}}{-3-\frac{1}{x^3}+\frac{2}{x^2}}=\frac{-2+0+0}{-3-0+0}$

$\therefore \lim_{x \to \infty} \frac{-2x^3+7x+2}{-3x^3-x+2}=\frac{2}{3}$