The problem in the image is the following limit expression:

$\lim_{{x \to +\infty}} \frac{x - \sqrt{x^2 + x + 1}}{2x + \sqrt{4x^2 + x}}$

To solve this limit as $x \to +\infty$, we use asymptotic analysis and simplifying techniques. The key idea is to factor out terms that dominate as $x \to \infty$ and deal with the square roots.

Step-by-step solution:

1. Focus on the numerator:

$x - \sqrt{x^2 + x + 1}$

Factor out $x$ from the square root: $\sqrt{x^2 + x + 1} = x \sqrt{1 + \frac{1}{x} + \frac{1}{x^2}}$ As $x \to \infty$, the terms $\frac{1}{x}$ and $\frac{1}{x^2}$ become negligible, so: $\sqrt{x^2 + x + 1} \sim x \left( 1 + \frac{1}{2x} \right)$ Thus, the expression becomes: $x - x \left( 1 + \frac{1}{2x} \right) = x - x - \frac{1}{2} = -\frac{1}{2}$ So, the numerator approaches $-\frac{1}{2}$ as $x \to \infty$.

2. Focus on the denominator:

$2x + \sqrt{4x^2 + x}$ Again, factor out $x$ from the square root: $\sqrt{4x^2 + x} = 2x \sqrt{1 + \frac{1}{4x}}$ As $x \to \infty$, $\frac{1}{4x}$ becomes negligible, so: $\sqrt{4x^2 + x} \sim 2x \left( 1 + \frac{1}{8x} \right)$ Thus, the denominator becomes: $2x + 2x \left( 1 + \frac{1}{8x} \right) = 2x + 2x + \frac{1}{4} = 4x + \frac{1}{4}$ So, the denominator approaches $4x$ as $x \to \infty$.

3. Combine the results:

Now, the limit can be simplified: $\lim_{{x \to +\infty}} \frac{-\frac{1}{2}}{4x} = 0$

Final answer:

$\lim_{{x \to +\infty}} \frac{x - \sqrt{x^2 + x + 1}}{2x + \sqrt{4x^2 + x}} = 0$

Would you like more details on any part of the process?

Related questions:

1. How does the technique of factoring out dominant terms help in limits involving square roots?
2. What are the most common ways to simplify expressions involving large $x$?
3. Why do terms like $\frac{1}{x}$ and $\frac{1}{x^2}$ become negligible as $x \to \infty$?
4. Can this limit be solved using L'Hôpital's Rule instead? How?
5. What are the potential pitfalls of not simplifying correctly in limits with square roots?

Tip:

When dealing with limits as $x \to \infty$, always consider the highest-order terms first, as lower-order terms often become negligible.