To solve the problem, we need to determine [tex]\(\lim_{n \to \infty} \frac{2a_n + b_n}{c_n}\)[/tex], given the limits of the sequences [tex]\(a_n\)[/tex], [tex]\(b_n\)[/tex], and [tex]\(c_n\)[/tex] as [tex]\(n\)[/tex] approaches infinity.
Given:
[tex]\[
\lim_{n \to \infty} a_n = 3,
\lim_{n \to \infty} b_n = -3,
\text{and}
\lim_{n \to \infty} c_n = \frac{1}{3}
\][/tex]
Step-by-step breakdown:
1. Evaluate the limit of the numerator:
[tex]\[
\lim_{n \to \infty} (2a_n + b_n)
\][/tex]
According to the given limits:
[tex]\[
\lim_{n \to \infty} (2a_n + b_n) = 2 \cdot \lim_{n \to \infty} a_n + \lim_{n \to \infty} b_n
\][/tex]
Substituting the given limits:
[tex]\[
= 2 \cdot 3 + (-3) = 6 - 3 = 3
\][/tex]
2. Evaluate the limit of the denominator:
[tex]\[
\lim_{n \to \infty} c_n = \frac{1}{3}
\][/tex]
3. Combine the results to find the limit of the ratio:
[tex]\[
\lim_{n \to \infty} \frac{2a_n + b_n}{c_n} = \frac{\lim_{n \to \infty} (2a_n + b_n)}{\lim_{n \to \infty} c_n}
\][/tex]
Substituting the calculated limits:
[tex]\[
= \frac{3}{\frac{1}{3}} = 3 \cdot 3 = 9
\][/tex]
Hence, the value of [tex]\(\lim_{n \to \infty} \frac{2a_n + b_n}{c_n}\)[/tex] is [tex]\(9\)[/tex].
So the correct answer is:
[tex]\[
\boxed{9}
\][/tex]
