To prove the given limit:

$\lim_{n \to \infty} n^2 \int_{0}^{\frac{1}{n}} x^{x+1} \, dx = \frac{1}{2},$

Step-by-step Solution:

Step 1: Approximation of $x^{x+1}$

For small $x$, we can approximate $x^{x+1}$ using properties of logarithms and exponential functions. We write: $x^{x+1} = x \cdot x^x.$ For $x^x$, we use: $x^x = e^{x \ln x}.$ When $x$ is very small, $x \ln x \to 0$, so $e^{x \ln x} \to 1$. Thus, for small $x$, $x^{x+1} \approx x.$

Step 2: Integral Approximation

The integral becomes: $\int_{0}^{\frac{1}{n}} x^{x+1} \, dx \approx \int_{0}^{\frac{1}{n}} x \, dx,$ because $x^{x+1} \approx x$. The integral of $x$ is: $\int_{0}^{\frac{1}{n}} x \, dx = \left[\frac{x^2}{2}\right]_{0}^{\frac{1}{n}} = \frac{\left(\frac{1}{n}\right)^2}{2} = \frac{1}{2n^2}.$

Step 3: Multiply by $n^2$

Now multiply the result of the integral by $n^2$: $n^2 \cdot \int_{0}^{\frac{1}{n}} x \, dx = n^2 \cdot \frac{1}{2n^2} = \frac{1}{2}.$

Step 4: Limit as $n \to \infty$

Since the integral approximation becomes exact as $n \to \infty$, the limit is: $\lim_{n \to \infty} n^2 \int_{0}^{\frac{1}{n}} x^{x+1} \, dx = \frac{1}{2}.$

Conclusion:

$\boxed{\lim_{n \to \infty} n^2 \int_{0}^{\frac{1}{n}} x^{x+1} \, dx = \frac{1}{2}}$

Let me know if you'd like a more detailed explanation or further clarification!

Expanding Questions:

1. Why is $x^{x+1} \approx x$ a valid approximation for small $x$?
2. How does the substitution $x \ln x$ simplify $x^x$?
3. Can we generalize this proof for different powers of $x$ (e.g., $x^{x+k}$)?
4. What is the impact of higher-order terms of $x^{x+1}$ on this integral?
5. How can this problem be visualized geometrically?

Tip:

Always check if approximations hold rigorously by bounding error terms when working with limits involving small $x$ or large $n$.