Express the following the limit values in terms of a definite integral and find them.

(1) $I=\lim_{n\to\infty}\frac{1}{n}\ln \left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)\cdots\cdots \left(1+\frac{n}{n}\right).$

(2) $J=\lim_{n\to\infty}\frac{1}{n^{2}}(\sqrt{n^{2}-1}+\sqrt{n^{2}-2^{2}}+\cdots\cdots+\sqrt{n^{2}-n^{2}}).$

(3) $K=\lim_{n\to\infty}\frac{1}{n^{3}}(\sqrt{n^{2}+1}+2\sqrt{n^{2}+2^{2}}+\cdots+n\sqrt{n^{2}+n^{2}}).$
1. We start with the limit expression for \( I \):

\[ I = \lim_{n \to \infty} \frac{1}{n} \ln \left( \left(1 + \frac{1}{n}\right) \left(1 + \frac{2}{n}\right) \cdots \left(1 + \frac{n}{n}\right) \right) \]

2. We can rewrite the product inside the logarithm as a sum of logarithms:

\[ I = \lim_{n \to \infty} \frac{1}{n} \ln \left( \prod_{k=1}^{n} \left(1 + \frac{k}{n}\right) \right) = \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \ln \left(1 + \frac{k}{n}\right) \]

3. Recognize that the sum inside the limit is a Riemann sum for the integral of \(\ln(1 + x)\) over the interval \([0, 1]\):

\[ I = \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \ln \left(1 + \frac{k}{n}\right) \approx \int_{0}^{1} \ln(1 + x) \, dx \]

4. Evaluate the integral:

\[ \int_{0}^{1} \ln(1 + x) \, dx \]

5. Use integration by parts, where \( u = \ln(1 + x) \) and \( dv = dx \):

\[ du = \frac{1}{1 + x} \, dx \quad \text{and} \quad v = x \]

6. Applying integration by parts:

\[ \int \ln(1 + x) \, dx = x \ln(1 + x) - \int \frac{x}{1 + x} \, dx \]

7. Simplify the integral:

\[ \int \frac{x}{1 + x} \, dx = \int \left(1 - \frac{1}{1 + x}\right) \, dx = x - \ln(1 + x) \]

8. Combine the results:

\[ \int \ln(1 + x) \, dx = x \ln(1 + x) - (x - \ln(1 + x)) = x \ln(1 + x) - x + \ln(1 + x) \]

9. Evaluate the definite integral from 0 to 1:

\[ \left[ x \ln(1 + x) - x + \ln(1 + x) \right]_{0}^{1} = \left[ 1 \ln(2) - 1 + \ln(2) \right] - \left[ 0 \ln(1) - 0 + \ln(1) \right] = 2 \ln(2) - 1 \]

10. Therefore, the value of \( I \) is:

\[ I = 2 \ln(2) - 1 \]

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1. For the second limit \( J \):

\[ J = \lim_{n \to \infty} \frac{1}{n^{2}} \left( \sqrt{n^{2} - 1} + \sqrt{n^{2} - 2^{2}} + \cdots + \sqrt{n^{2} - n^{2}} \right) \]

2. Rewrite the sum inside the limit:

\[ J = \lim_{n \to \infty} \frac{1}{n^{2}} \sum_{k=1}^{n} \sqrt{n^{2} - k^{2}} \]

3. Factor out \( n^{2} \) from the square root:

\[ J = \lim_{n \to \infty} \frac{1}{n^{2}} \sum_{k=1}^{n} n \sqrt{1 - \left(\frac{k}{n}\right)^{2}} = \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \sqrt{1 - \left(\frac{k}{n}\right)^{2}} \]

4. Recognize that the sum inside the limit is a Riemann sum for the integral of \(\sqrt{1 - x^{2}}\) over the interval \([0, 1]\):

\[ J = \int_{0}^{1} \sqrt{1 - x^{2}} \, dx \]

5. This integral represents the area of a quarter circle of radius 1:

\[ \int_{0}^{1} \sqrt{1 - x^{2}} \, dx = \frac{\pi}{4} \]

6. Therefore, the value of \( J \) is:

\[ J = \frac{\pi}{4} \]

The final answer is \( \boxed{ I = 2 \ln(2) - 1 } \) and \( J = \frac{\pi}{4} \)