We are tasked with finding the limit:

$\lim_{x \to \infty} \sqrt{\frac{8x^2 - 3}{2x^2 + x}}$

Step 1: Simplify the expression inside the square root.

The expression inside the square root is:

$\frac{8x^2 - 3}{2x^2 + x}$

To handle limits as $x \to \infty$, it helps to factor out the highest power of $x$ from both the numerator and the denominator. In this case, $x^2$ is the highest power.

$\frac{8x^2 - 3}{2x^2 + x} = \frac{x^2(8 - \frac{3}{x^2})}{x^2(2 + \frac{1}{x})}$

Now, we can cancel out $x^2$ from both the numerator and the denominator:

$= \frac{8 - \frac{3}{x^2}}{2 + \frac{1}{x}}$

Step 2: Take the limit as $x \to \infty$.

As $x \to \infty$, the terms $\frac{3}{x^2}$ and $\frac{1}{x}$ approach 0. Therefore, the expression simplifies to:

$\frac{8 - 0}{2 + 0} = \frac{8}{2} = 4$

Step 3: Apply the square root.

Now, apply the square root to the result:

$\sqrt{4} = 2$

Final Answer:

$\lim_{x \to \infty} \sqrt{\frac{8x^2 - 3}{2x^2 + x}} = 2$

Would you like further details or explanations on any of the steps?

Here are 5 related questions you might explore:

1. How would the limit change if the coefficient of $x^2$ in the numerator were different?
2. What is the effect of the lower degree terms like $x$ and constants on limits at infinity?
3. How do square roots interact with limits in general?
4. What if $x \to -\infty$ instead of $x \to \infty$ for this limit?
5. How do you handle limits of more complicated expressions involving polynomials?

Tip: When calculating limits at infinity, focus on the highest degree terms, as they dominate the behavior of the function.