Question

determine the following limit. (limlimits_{x \to -infty} \frac{19x^3 + 8x^2 - 8x}{15x^3 + 7x^2 + 7x + 1}) select the correct choice and, if necessary, fill in the answer box to complete your choice. (\bigcirc) a. (limlimits_{x \to -infty} \frac{19x^3 + 8x^2 - 8x}{15x^3 + 7x^2 + 7x + 1} = square) (simplify your answer.) (\bigcirc) b. the limit does not exist and is neither (-infty) nor (infty).

Explanation:

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

For the numerator: \(\frac{19x^3 + 8x^2 - 8x}{x^3}=19+\frac{8}{x}-\frac{8}{x^2}\)
For the denominator: \(\frac{15x^3 + 7x^2 + 7x + 1}{x^3}=15+\frac{7}{x}+\frac{7}{x^2}+\frac{1}{x^3}\)

Step2: Take the limit as \(x\to -\infty\)

As \(x\to -\infty\), \(\frac{1}{x}\to 0\), \(\frac{1}{x^2}\to 0\), \(\frac{1}{x^3}\to 0\).
So, \(\lim_{x\to -\infty}\frac{19+\frac{8}{x}-\frac{8}{x^2}}{15+\frac{7}{x}+\frac{7}{x^2}+\frac{1}{x^3}}=\frac{19 + 0 - 0}{15 + 0 + 0 + 0}=\frac{19}{15}\)

Answer:

A. \(\lim\limits_{x\to -\infty}\frac{19x^3 + 8x^2 - 8x}{15x^3 + 7x^2 + 7x + 1}=\frac{19}{15}\)