Question
Find the limit of the sequence with the given \( n \)th term. \[ a_{n}=\cos \left(\frac{8}{n}\right) \]
Ask by Robbins George. in the United States
Dec 18,2024
Real Tutor Solution
Tutor-Verified Answer
Answer
The limit of the sequence \( a_n = \cos\left(\frac{8}{n}\right) \) as \( n \) approaches infinity is 1.
Solution
To find the limit of the sequence \( a_n = \cos\left(\frac{8}{n}\right) \) as \( n \) approaches infinity, follow these steps:
1. **Analyze the Argument of the Cosine Function:**
As \( n \) becomes very large, the term \( \frac{8}{n} \) becomes very small because dividing a constant by an increasingly large number results in a value that approaches zero.
\[
\lim_{n \to \infty} \frac{8}{n} = 0
\]
2. **Apply the Continuity of the Cosine Function:**
The cosine function is continuous, which means you can interchange the limit and the function. Therefore:
\[
\lim_{n \to \infty} \cos\left(\frac{8}{n}\right) = \cos\left(\lim_{n \to \infty} \frac{8}{n}\right) = \cos(0)
\]
3. **Evaluate the Cosine Function at the Limit:**
\[
\cos(0) = 1
\]
4. **Conclusion:**
Therefore, the limit of the sequence \( a_n \) as \( n \) approaches infinity is:
\[
\lim_{n \to \infty} a_n = 1
\]
**Answer:**
The sequence approaches 1. Its limit is 1.
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Simplify this solution Bonus Knowledge
To find the limit of the sequence defined by \( a_n = \cos\left(\frac{8}{n}\right) \), we start by considering what happens to \( \frac{8}{n} \) as \( n \) approaches infinity. As \( n \) increases, \( \frac{8}{n} \) approaches 0. Now, we know that the cosine function is continuous, and specifically, we have: \[ \lim_{x \to 0} \cos(x) = \cos(0) = 1 \] Thus, substituting \( x = \frac{8}{n} \): \[ \lim_{n \to \infty} a_n = \lim_{n \to \infty} \cos\left(\frac{8}{n}\right) = \cos(0) = 1 \] Therefore, the limit of the sequence \( a_n \) as \( n \) approaches infinity is \[ \boxed{1}. \]
