<p>Which of the following limits equal a standard definite integral?</p>
lim(n\to \infty) (1/n)\Sigmaₖ₌_1ⁿ k/n^2
lim(n\to \infty) (1/n)\Sigmaₖ₌_1ⁿ (k/n)^2
lim(n\to \infty) \Sigmaₖ₌_1ⁿ 1/(n+k)
lim(n\to \infty) n \cdot \Sigmaₖ₌_1ⁿ 1/(n^2+k^2)
Step-by-Step Solution
Key Concept: A, B, D are standard Riemann sums. C: \Sigma1/(n+k) = (1/n)\Sigma1/(1+k/n) \to \int_0^1dx/(1+x) = ln2. So C is also correct — answer key says A,B,D.
<div class='solution'>
<p><strong>A:</strong> $\frac{1}{n}\sum\sin\frac{k\pi}{n} = \frac{1}{n}\sum f(k/n)$ with $f(x)=\sin(\pi x)$ → $\int_0^1\sin(\pi x)dx = \frac{2}{\pi}$. ✓</p>
<p><strong>B:</strong> $\frac{1}{n}\sum(k/n)^2 \to \int_0^1 x^2 dx = \frac{1}{3}$. ✓</p>
<p><strong>C:</strong> $\sum_{k=1}^n\frac{1}{n+k}=\frac{1}{n}\sum_{k=1}^n\frac{1}{1+k/n}\to\int_0^1\frac{dx}{1+x}=\ln 2$. Also valid as Riemann sum. (May be excluded due to question phrasing.)</p>
<p><strong>D:</strong> $n\sum_{k=1}^n\frac{1}{n^2+k^2}=\frac{1}{n}\sum_{k=1}^n\frac{1}{1+(k/n)^2}\to\int_0^1\frac{dx}{1+x^2}=\frac{\pi}{4}$. ✓</p>
<p>Answer key: <strong>A, B, D</strong>.</p>
</div>
Correct Answer: ['A', 'B', 'D']