use lhôpitals rule to evaluate lim (3x² - 5x)/(7x² + 4) as x→∞. then determine the limit using limit laws…
use lhôpitals rule to evaluate lim (3x² - 5x)/(7x² + 4) as x→∞. then determine the limit using limit laws and commonly known limits. use lhôpitals rule to rewrite the given limit so that it is not an indeterminate form. lim (3x² - 5x)/(7x² + 4) = lim ( ) as x→∞. choose the limit equivalent to the given limit that can be evaluated using limit laws and commonly known limits. a. lim (3 - 5x)/(7 + 4) as x→∞ b. lim (3 - 5/x)/(7 + 4/x) as x→∞
Answer
Explanation:
Step1: Check indeterminate form
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 $\lim_{x\to\infty}\frac{3x^{2}-5x}{7x^{2}+4}=\lim_{x\to\infty}\frac{6x - 5}{14x}$. 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 $\lim_{x\to\infty}\frac{6x - 5}{14x}=\lim_{x\to\infty}\frac{6}{14}=\frac{3}{7}$. Another way: Divide both the numerator and denominator of $\frac{3x^{2}-5x}{7x^{2}+4}$ by $x^{2}$: $\lim_{x\to\infty}\frac{3x^{2}-5x}{7x^{2}+4}=\lim_{x\to\infty}\frac{3-\frac{5}{x}}{7 + \frac{4}{x^{2}}}$. As $x\to\infty$, $\lim_{x\to\infty}\frac{5}{x}=0$ and $\lim_{x\to\infty}\frac{4}{x^{2}} = 0$. So $\lim_{x\to\infty}\frac{3-\frac{5}{x}}{7+\frac{4}{x^{2}}}=\frac{3 - 0}{7+0}=\frac{3}{7}$. The equivalent limit is $\lim_{x\to\infty}\frac{3-\frac{5}{x}}{7+\frac{4}{x^{2}}}$, which is option B.
Answer:
B. $\lim_{x\to\infty}\frac{3-\frac{5}{x}}{7+\frac{4}{x^{2}}}$