Hall Patel
12/21/2023 · Primary School
19. \( \lim _{n \rightarrow \infty} \frac{(n-1)(2 n+1)(2 n+1)(4 n-1)}{n^{4}} \)
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Step-by-step Solution
Calculate the limit \( \lim _{n \rightarrow \infty} \frac{(n-1)(2 n+1)(2 n+1)(4 n-1)}{n^{4}} \).
Evaluate the limit by following steps:
- step0: Evaluate using transformations:
\(\lim _{n\rightarrow +\infty}\left(\frac{\left(n-1\right)\left(2n+1\right)\left(2n+1\right)\left(4n-1\right)}{n^{4}}\right)\)
- step1: Multiply the first two terms:
\(\lim _{n\rightarrow +\infty}\left(\frac{\left(n-1\right)\left(2n+1\right)^{2}\left(4n-1\right)}{n^{4}}\right)\)
- step2: Rewrite the expression:
\(\lim _{n\rightarrow +\infty}\left(\frac{n^{4}}{n^{4}}\times \frac{16-\frac{4}{n}-\frac{12}{n^{2}}-\frac{1}{n^{3}}+\frac{1}{n^{4}}}{1}\right)\)
- step3: Simplify:
\(\lim _{n\rightarrow +\infty}\left(\frac{16-\frac{4}{n}-\frac{12}{n^{2}}-\frac{1}{n^{3}}+\frac{1}{n^{4}}}{1}\right)\)
- step4: Rewrite the expression:
\(\frac{\lim _{n\rightarrow +\infty}\left(16-\frac{4}{n}-\frac{12}{n^{2}}-\frac{1}{n^{3}}+\frac{1}{n^{4}}\right)}{1}\)
- step5: Rewrite the expression:
\(\frac{16}{1}\)
- step6: Divide the terms:
\(16\)
The limit of the given expression as \( n \) approaches infinity is 16.
Quick Answer
The limit is 16.
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