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

Explanation:

Step1: Apply L'Hopital's Rule

The original limit $\lim_{x\rightarrow\infty}\frac{3x^{2}-5x}{7x^{2}+4}$ is in the indeterminate form $\frac{\infty}{\infty}$. Differentiate the numerator and denominator. The derivative of $3x^{2}-5x$ is $6x - 5$ and the derivative of $7x^{2}+4$ is $14x$. So, $\lim_{x\rightarrow\infty}\frac{3x^{2}-5x}{7x^{2}+4}=\lim_{x\rightarrow\infty}\frac{6x - 5}{14x}$. This is still in the $\frac{\infty}{\infty}$ form. Apply L'Hopital's Rule again. The derivative of $6x - 5$ is $6$ and the derivative of $14x$ is $14$. So $\lim_{x\rightarrow\infty}\frac{6x - 5}{14x}=\lim_{x\rightarrow\infty}\frac{6}{14}=\frac{3}{7}$.

Step2: Use limit - laws

Divide both the numerator and denominator of $\frac{3x^{2}-5x}{7x^{2}+4}$ by $x^{2}$: $\lim_{x\rightarrow\infty}\frac{3x^{2}-5x}{7x^{2}+4}=\lim_{x\rightarrow\infty}\frac{3-\frac{5}{x}}{7 + \frac{4}{x^{2}}}$. We know that $\lim_{x\rightarrow\infty}\frac{1}{x}=0$ and $\lim_{x\rightarrow\infty}\frac{1}{x^{2}} = 0$. So, $\lim_{x\rightarrow\infty}\frac{3-\frac{5}{x}}{7+\frac{4}{x^{2}}}=\frac{3-0}{7 + 0}=\frac{3}{7}$.

Answer:

The equivalent limit using limit - laws is B. $\lim_{x\rightarrow\infty}\frac{3-\frac{5}{x}}{7+\frac{4}{x^{2}}}$, and the value of the limit $\lim_{x\rightarrow\infty}\frac{3x^{2}-5x}{7x^{2}+4}=\frac{3}{7}$