Question

use lhôpitals rule to find the following limit.

$$\lim_{x \to \infty} \frac{3x^3 - 6x}{5x^3 + 7}$$

$$\lim_{x \to \infty} \frac{3x^3 - 6x}{5x^3 + 7} = \square$$ (type an integer or a fraction.)

Explanation:

Step1: Check indeterminate form

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

Step2: Apply L'Hôpital's rule

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

Step3: Simplify the limit

Divide numerator and denominator by \(x^{2}\).
\(\lim_{x\to\infty}\frac{9x^{2}-6}{15x^{2}}=\lim_{x\to\infty}\frac{9-\frac{6}{x^{2}}}{15}\).
As \(x\to\infty\), \(\frac{6}{x^{2}}\to0\).

Answer:

\(\frac{3}{5}\)