Question

evaluate the following limit. use lhôpitals rule when it is convenient and applicable.
lim _ { x
ightarrow infty } \frac { 3 x ^ { 3 } - 6 x } { 5 x ^ { 3 } + 7 }
how should the given limit be evaluated? select the correct choice below and, if necessary, fill in the answer box to complete your choic
a. use lhôpitals rule exactly once to rewrite the limit as ( lim _ { x
ightarrow infty } ) ().
b. use lhôpitals rule more than once to rewrite the limit in its final form as ( lim _ { x
ightarrow infty } ) ().
c. use direct substitution.
d. multiply the expression by a unit fraction to obtain ( lim _ { x
ightarrow infty } ) ().

Explanation:

Step1: Apply L'Hôpital's Rule first time

Differentiate numerator and denominator.
The derivative of \(3x^{3}-6x\) is \(9x^{2}-6\), and the derivative of \(5x^{3}+7\) is \(15x^{2}\).
So the limit becomes \(\lim_{x
ightarrow\infty}\frac{9x^{2}-6}{15x^{2}}\)

Step2: Apply L'Hôpital's Rule second time

Differentiate numerator and denominator again.
The derivative of \(9x^{2}-6\) is \(18x\), and the derivative of \(15x^{2}\) is \(30x\).
So the limit becomes \(\lim_{x
ightarrow\infty}\frac{18x}{30x}\)

Step3: Simplify the limit

Cancel out \(x\) (since \(x
eq0\) as \(x
ightarrow\infty\)), we get \(\lim_{x
ightarrow\infty}\frac{18}{30}=\frac{3}{5}\)

Answer:

\(\frac{3}{5}\)