The problem is to calculate:

$\lim_{x \to +\infty} \sqrt{x^2 + x} - x.$

Solution:

1. Factor $x^2$ inside the square root: Inside the square root, factor out $x^2$: $\sqrt{x^2 + x} = \sqrt{x^2(1 + \frac{1}{x})} = x\sqrt{1 + \frac{1}{x}},$ since $x > 0$ as $x \to +\infty$.

2. Rewrite the expression: Substitute $x\sqrt{1 + \frac{1}{x}}$ into the limit: $\lim_{x \to +\infty} \sqrt{x^2 + x} - x = \lim_{x \to +\infty} x\sqrt{1 + \frac{1}{x}} - x.$

3. Factor $x$ out of the expression: Factor $x$ from the terms: $x\sqrt{1 + \frac{1}{x}} - x = x\left(\sqrt{1 + \frac{1}{x}} - 1\right).$

4. Simplify $\sqrt{1 + \frac{1}{x}} - 1$: Use the approximation $\sqrt{1 + u} \approx 1 + \frac{u}{2}$ for small $u$ (here $u = \frac{1}{x}$): $\sqrt{1 + \frac{1}{x}} \approx 1 + \frac{1}{2x}.$ So: $\sqrt{1 + \frac{1}{x}} - 1 \approx \frac{1}{2x}.$

5. Substitute the approximation back: Substitute $\sqrt{1 + \frac{1}{x}} - 1 \approx \frac{1}{2x}$ into the expression: $x\left(\sqrt{1 + \frac{1}{x}} - 1\right) \approx x \cdot \frac{1}{2x} = \frac{1}{2}.$

6. Conclusion: The limit is: $\lim_{x \to +\infty} \sqrt{x^2 + x} - x = \frac{1}{2}.$

Final Answer:

$\boxed{\frac{1}{2}}$

Let me know if you'd like further clarifications! Here are five related questions to deepen your understanding:

Related Questions:

1. How would the limit change if the expression were $\sqrt{x^2 + 2x} - x$?
2. What is the behavior of $\sqrt{x^2 + x} + x$ as $x \to +\infty$?
3. How can you rigorously justify the approximation $\sqrt{1 + u} \approx 1 + \frac{u}{2}$ using a binomial expansion?
4. What happens if $x$ approaches $-\infty$ instead of $+\infty$?
5. Can the same method be used for $\sqrt{x^2 + ax} - bx$ for constants $a$ and $b$? Why or why not?

Tip:

When simplifying expressions with square roots, factoring out the dominant term (e.g., $x^2$) is a reliable strategy to isolate smaller terms and approximate the behavior of the function at infinity.