Question

evaluate the following limit. use lhôpitals rule when it is convenient and applicable.
limₓ→∞ (3x³ - 6x)/(5x³ + 7)
how should the given limit be evaluated? select the correct choice below and, if necessary, fill in the answer box to complete
○ a. use lhôpitals rule exactly once to rewrite the limit as limₓ→∞ (□).
○ b. use lhôpitals rule more than once to rewrite the limit in its final form as limₓ→∞ (□).
○ c. use direct substitution.
○ d. multiply the expression by a unit fraction to obtain limₓ→∞ (□).

Explanation:

Step1: Check the form of the limit

When \(x\to\infty\), \(\lim_{x\to\infty}\frac{3x^{3}-6x}{5x^{3}+7}\) is in the \(\frac{\infty}{\infty}\) form.

Step2: Apply L'Hôpital's Rule once

Differentiate the numerator and denominator. The derivative of \(y = 3x^{3}-6x\) is \(y^\prime=9x^{2}-6\), and the derivative of \(y = 5x^{3}+7\) is \(y^\prime = 15x^{2}\). So, by L'Hôpital's Rule, \(\lim_{x\to\infty}\frac{3x^{3}-6x}{5x^{3}+7}=\lim_{x\to\infty}\frac{9x^{2}-6}{15x^{2}}\).

Answer:

A. Use l'Hôpital's Rule exactly once to rewrite the limit as \(\lim_{x\to\infty}\frac{9x^{2}-6}{15x^{2}}\)