Question

use lhôpitals rule to evaluate \\( \lim _{x \
ightarrow \infty} \frac{5 x^{2}-7 x}{9 x^{2}+4} \\). 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}{9 x^{2}+4}=\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 \(g(x)=9x^{2}+4\), \(g^{\prime}(x)=18x\).
So, \(\lim_{x
ightarrow\infty}\frac{5x^{2}-7x}{9x^{2}+4}=\lim_{x
ightarrow\infty}\frac{10x - 7}{18x}\).

Step2: Check the form and apply L'Hôpital's Rule again (if needed)

The limit \(\lim_{x
ightarrow\infty}\frac{10x - 7}{18x}\) is still in the \(\frac{\infty}{\infty}\) form.
Differentiate \(f_1(x)=10x - 7\) (so \(f_1^{\prime}(x)=10\)) and \(g_1(x)=18x\) (so \(g_1^{\prime}(x)=18\)).
Then \(\lim_{x
ightarrow\infty}\frac{10x - 7}{18x}=\lim_{x
ightarrow\infty}\frac{10}{18}\).

Step3: Use limit laws (alternative method)

Divide numerator and denominator by \(x^{2}\):
\(\lim_{x
ightarrow\infty}\frac{5x^{2}-7x}{9x^{2}+4}=\lim_{x
ightarrow\infty}\frac{5-\frac{7}{x}}{9+\frac{4}{x^{2}}}\).
Using the limit laws \(\lim_{x
ightarrow\infty}\frac{1}{x}=0\) and \(\lim_{x
ightarrow\infty}\frac{1}{x^{2}} = 0\).
\(\lim_{x
ightarrow\infty}\frac{5-\frac{7}{x}}{9+\frac{4}{x^{2}}}=\frac{\lim_{x
ightarrow\infty}(5 - 7\lim_{x
ightarrow\infty}\frac{1}{x})}{\lim_{x
ightarrow\infty}(9+4\lim_{x
ightarrow\infty}\frac{1}{x^{2}})}\).

Answer:

\(\frac{5}{9}\)