Question
Find the limit of the sequence: \( a_{n}=\frac{4 n-2}{9 n+1} \)
Ask by Klein Nguyen. in the United States
Mar 11,2025
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Answer
The limit of the sequence \( a_{n}=\frac{4n-2}{9n+1} \) as \( n \) approaches infinity is \( \frac{4}{9} \) or \( 0.\dot{4} \).
Solution
Calculate the limit \( \lim_{n\to\infty} \frac{4n-2}{9n+1} \).
Evaluate the limit by following steps:
- step0: Evaluate using transformations:
\(\lim _{n\rightarrow +\infty}\left(\frac{4n-2}{9n+1}\right)\)
- step1: Rewrite the expression:
\(\lim _{n\rightarrow +\infty}\left(\frac{n\left(4-\frac{2}{n}\right)}{n\left(9+\frac{1}{n}\right)}\right)\)
- step2: Reduce the fraction:
\(\lim _{n\rightarrow +\infty}\left(\frac{4-\frac{2}{n}}{9+\frac{1}{n}}\right)\)
- step3: Rewrite the expression:
\(\frac{\lim _{n\rightarrow +\infty}\left(4-\frac{2}{n}\right)}{\lim _{n\rightarrow +\infty}\left(9+\frac{1}{n}\right)}\)
- step4: Calculate:
\(\frac{4}{\lim _{n\rightarrow +\infty}\left(9+\frac{1}{n}\right)}\)
- step5: Calculate:
\(\frac{4}{9}\)
The limit of the sequence \( a_{n}=\frac{4 n-2}{9 n+1} \) as \( n \) approaches infinity is \( \frac{4}{9} \) or \( 0.\dot{4} \).
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Simplify this solution Bonus Knowledge
To find the limit of the sequence \( a_{n}=\frac{4 n-2}{9 n+1} \) as \( n \) approaches infinity, we can analyze the expression by dividing the numerator and the denominator by \( n \): \[ a_{n} = \frac{4 - \frac{2}{n}}{9 + \frac{1}{n}} \] As \( n \) approaches infinity, the terms \( \frac{2}{n} \) and \( \frac{1}{n} \) both approach \( 0 \). Therefore, we have: \[ \lim_{n \to \infty} a_{n} = \frac{4 - 0}{9 + 0} = \frac{4}{9} \] Thus, the limit of the sequence is \( \frac{4}{9} \).
