Question

use lhôpitals rule to evaluate \\( \lim _ { x \
ightarrow \infty } \frac { 2 x ^ { 2 } + 3 x } { 4 x ^ { 3 } + 3 x + 1 } \\). 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 { 2 x ^ { 2 } + 3 x } { 4 x ^ { 3 } + 3 x + 1 } = \lim _ { x \
ightarrow \infty } \\)

Explanation:

Step1: Apply L'Hôpital's Rule once

Differentiate numerator and denominator.
The derivative of \(2x^{2}+3x\) is \(4x + 3\), and the derivative of \(4x^{3}+3x + 1\) is \(12x^{2}+3\).
So, \(\lim_{x
ightarrow\infty}\frac{2x^{2}+3x}{4x^{3}+3x + 1}=\lim_{x
ightarrow\infty}\frac{4x + 3}{12x^{2}+3}\)

Step2: Apply L'Hôpital's Rule again

Differentiate numerator and denominator again.
The derivative of \(4x + 3\) is \(4\), and the derivative of \(12x^{2}+3\) is \(24x\).
So, \(\lim_{x
ightarrow\infty}\frac{4x + 3}{12x^{2}+3}=\lim_{x
ightarrow\infty}\frac{4}{24x}\)

Step3: Evaluate the limit

As \(x
ightarrow\infty\), \(\lim_{x
ightarrow\infty}\frac{4}{24x}=0\)

Now, using limit laws:
Divide numerator and denominator by \(x^{3}\):
\(\lim_{x
ightarrow\infty}\frac{2x^{2}+3x}{4x^{3}+3x + 1}=\lim_{x
ightarrow\infty}\frac{\frac{2x^{2}}{x^{3}}+\frac{3x}{x^{3}}}{\frac{4x^{3}}{x^{3}}+\frac{3x}{x^{3}}+\frac{1}{x^{3}}}=\lim_{x
ightarrow\infty}\frac{\frac{2}{x}+\frac{3}{x^{2}}}{4+\frac{3}{x^{2}}+\frac{1}{x^{3}}}\)
As \(x
ightarrow\infty\), \(\frac{2}{x}
ightarrow0\), \(\frac{3}{x^{2}}
ightarrow0\), \(\frac{1}{x^{3}}
ightarrow0\)
So, \(\lim_{x
ightarrow\infty}\frac{\frac{2}{x}+\frac{3}{x^{2}}}{4+\frac{3}{x^{2}}+\frac{1}{x^{3}}}=0\)

Answer:

\(\lim_{x
ightarrow\infty}\frac{2x^{2}+3x}{4x^{3}+3x + 1}=0\)