Question

use lhôpitals rule to evaluate \\( \lim _{x \
ightarrow \infty} \frac{5 x^{2}-7 x}{7 x^{2}+6} \\). 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 _{x \
ightarrow \infty} \frac{5 x^{2}-7 x}{7 x^{2}+6}=\lim _{x \
ightarrow \infty}(\square) \\)

Explanation:

Step1: Apply L'Hôpital's Rule

L'Hôpital's Rule states that if \(\lim_{x
ightarrow a}\frac{f(x)}{g(x)}\) is in the indeterminate form \(\frac{\infty}{\infty}\) or \(\frac{0}{0}\), then \(\lim_{x
ightarrow a}\frac{f(x)}{g(x)}=\lim_{x
ightarrow a}\frac{f^{\prime}(x)}{g^{\prime}(x)}\).
For \(f(x) = 5x^{2}-7x\), \(f^{\prime}(x)=10x - 7\).
For \(g(x)=7x^{2}+6\), \(g^{\prime}(x)=14x\).
So, \(\lim_{x
ightarrow\infty}\frac{5x^{2}-7x}{7x^{2}+6}=\lim_{x
ightarrow\infty}\frac{10x - 7}{14x}\).

Step2: Apply L'Hôpital's Rule again (if needed)

The limit \(\lim_{x
ightarrow\infty}\frac{10x - 7}{14x}\) is still in the \(\frac{\infty}{\infty}\) form.
Differentiate \(f_1(x)=10x - 7\) (so \(f_1^{\prime}(x)=10\)) and \(g_1(x)=14x\) (so \(g_1^{\prime}(x)=14\)).
Then \(\lim_{x
ightarrow\infty}\frac{10x - 7}{14x}=\lim_{x
ightarrow\infty}\frac{10}{14}\).

Answer:

\(\frac{5}{7}\)