Evaluate the limits in Exercise 9 and 10 by identifying them with definite integrals and evaluating the integrals.
lim (n → ∞) Σ (from k=1 to n) ln √(1 + k/n)

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Recognize that the given limit is a Riemann sum of the form \(\lim_{n \to \infty} \sum_{k=1}^n f\left(\frac{k}{n}\right) \Delta x\), where \(\Delta x = \frac{1}{n}\) and \(f(x) = \ln \sqrt{1 + x}\).

Rewrite the summand \(\ln \sqrt{1 + \frac{k}{n}}\) as \(\frac{1}{2} \ln \left(1 + \frac{k}{n}\right)\) to simplify the expression inside the sum.

Express the sum as \(\sum_{k=1}^n \frac{1}{2} \ln \left(1 + \frac{k}{n}\right) \cdot \frac{1}{n}\), which matches the form of a Riemann sum for the integral of \(\frac{1}{2} \ln(1 + x)\) over the interval \([0,1]\).

Set up the definite integral corresponding to the limit: \(\int_0^1 \frac{1}{2} \ln(1 + x) \, dx\).

Evaluate the integral \(\int_0^1 \frac{1}{2} \ln(1 + x) \, dx\) using integration techniques such as integration by parts to find the exact value representing the limit.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Definite Integral as a Limit of Riemann Sums

A definite integral can be interpreted as the limit of a sum of function values times small intervals, known as a Riemann sum. As the number of subintervals increases indefinitely, the sum approaches the exact area under the curve. Recognizing a limit of sums as a definite integral is key to evaluating such limits.

Properties of Logarithmic Functions

Logarithmic functions, such as ln(x), have properties that simplify expressions, like ln(√x) = (1/2)ln(x). Understanding these properties helps in rewriting the sum's terms into a more manageable form, facilitating the identification of the integrand in the limit process.

Limit of a Sequence and Continuity

Evaluating limits involving sequences often requires understanding how functions behave as their inputs approach a point. Continuity ensures that the limit of the function values equals the function value at the limit point, allowing substitution inside the integral after identifying the integrand.

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