Question:
The value of \(\lim_{n \to \infty} \left\{\frac{1}{na} + \frac{1}{na+1} + \frac{1}{na+2} + \ldots + \frac{1}{nb}\right\}\) is
The value of \(\lim_{n \to \infty} \left\{\frac{1}{na} + \frac{1}{na+1} + \frac{1}{na+2} + \ldots + \frac{1}{nb}\right\}\) is
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Recognize as a Riemann sum: \(\lim_{n \to \infty} \frac{1}{n} \sum f(k/n) = \int f(x)dx\).
Updated On: Apr 23, 2026
- \(\log(ab)\)
- \(\log(a/b)\)
- \(\log(b/a)\)
- \(\log(a+b)\)
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The Correct Option is C
Solution and Explanation
Step 1: Formula / Definition}
\[ \lim_{n \to \infty} \frac{1}{n} \sum_{k=0}^{n(b-a)} \frac{1}{a + k/n} \]
Step 2: Calculation / Simplification}
\(= \int_0^{b-a} \frac{dx}{a+x} = [\log(a+x)]_0^{b-a}\)
\(= \log(a+b-a) - \log a = \log b - \log a = \log(b/a)\)
Step 3: Final Answer
\[ \log(b/a) \]
\[ \lim_{n \to \infty} \frac{1}{n} \sum_{k=0}^{n(b-a)} \frac{1}{a + k/n} \]
Step 2: Calculation / Simplification}
\(= \int_0^{b-a} \frac{dx}{a+x} = [\log(a+x)]_0^{b-a}\)
\(= \log(a+b-a) - \log a = \log b - \log a = \log(b/a)\)
Step 3: Final Answer
\[ \log(b/a) \]
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