Question

use lhôpitals rule to find the following limit.

$$\lim_{x\to\infty}\frac{2x^{3}-6x}{3x^{3}+7}$$

$$\lim_{x\to\infty}\frac{2x^{3}-6x}{3x^{3}+7}=\square$$ (type an integer or a fraction.)

Explanation:

Step1: Check the form of the limit

As \(x\to\infty\), \(\lim_{x\to\infty}\frac{2x^{3}-6x}{3x^{3}+7}\) is in the \(\frac{\infty}{\infty}\) form. So, we can apply L'Hôpital's rule.
Differentiate the numerator and denominator.
The derivative of \(y = 2x^{3}-6x\) is \(y^\prime=6x^{2}-6\) (using the power rule \((x^n)^\prime = nx^{n - 1}\)).
The derivative of \(y = 3x^{3}+7\) is \(y^\prime = 9x^{2}\).
So, \(\lim_{x\to\infty}\frac{2x^{3}-6x}{3x^{3}+7}=\lim_{x\to\infty}\frac{6x^{2}-6}{9x^{2}}\).

Step2: Simplify the new limit

\(\lim_{x\to\infty}\frac{6x^{2}-6}{9x^{2}}=\lim_{x\to\infty}(\frac{6x^{2}}{9x^{2}}-\frac{6}{9x^{2}})\).
Since \(\lim_{x\to\infty}\frac{1}{x^{2}} = 0\), we have \(\lim_{x\to\infty}\frac{6x^{2}}{9x^{2}}-\lim_{x\to\infty}\frac{6}{9x^{2}}\).
\(\lim_{x\to\infty}\frac{6x^{2}}{9x^{2}}=\frac{6}{9}=\frac{2}{3}\) and \(\lim_{x\to\infty}\frac{6}{9x^{2}} = 0\).

Answer:

\(\frac{2}{3}\)