use lhopitals rule to evaluate lim x→∞ 3x² - 5x / 7x² + 4. then determine the limit using limit laws and…

Explanation:

Step1: Check form for L'Hopital's Rule

As $x\to\infty$, $\lim_{x\to\infty}\frac{3x^{2}-5x}{7x^{2}+4}$ is in the $\frac{\infty}{\infty}$ form.

Step2: Apply L'Hopital's Rule

Differentiate the numerator and denominator. The derivative of $3x^{2}-5x$ is $6x - 5$, and the derivative of $7x^{2}+4$ is $14x$. So we get $\lim_{x\to\infty}\frac{6x - 5}{14x}$. This is still in the $\frac{\infty}{\infty}$ form.

Step3: Apply L'Hopital's Rule again

Differentiate the new - numerator and denominator. The derivative of $6x - 5$ is $6$, and the derivative of $14x$ is $14$. So $\lim_{x\to\infty}\frac{6x - 5}{14x}=\lim_{x\to\infty}\frac{6}{14}=\frac{3}{7}$.

Step4: Use limit laws

Divide both numerator and denominator of $\frac{3x^{2}-5x}{7x^{2}+4}$ by $x^{2}$: $\lim_{x\to\infty}\frac{3x^{2}-5x}{7x^{2}+4}=\lim_{x\to\infty}\frac{3-\frac{5}{x}}{7 + \frac{4}{x^{2}}}$. Since $\lim_{x\to\infty}\frac{1}{x}=0$ and $\lim_{x\to\infty}\frac{1}{x^{2}} = 0$, we have $\lim_{x\to\infty}\frac{3-\frac{5}{x}}{7+\frac{4}{x^{2}}}=\frac{3-0}{7 + 0}=\frac{3}{7}$.

Answer:

$\frac{3}{7}$