We are given the following limit problem:

$\lim_{{x \to \infty}} \frac{\sqrt{8x^2 + 5x - 7}}{\sqrt{2x^2 + 3x - 1}}$

Step-by-step solution:

1. Simplifying the square roots: Since both the numerator and the denominator involve square roots of quadratic expressions, we can factor out $x^2$ from inside the square roots.

   • In the numerator: $\sqrt{8x^2 + 5x - 7} = \sqrt{x^2 \left( 8 + \frac{5}{x} - \frac{7}{x^2} \right)} = |x| \sqrt{8 + \frac{5}{x} - \frac{7}{x^2}}$ As $x \to \infty$, we can drop the absolute value of $x$ because $x$ is positive in this limit.

   • In the denominator: $\sqrt{2x^2 + 3x - 1} = \sqrt{x^2 \left( 2 + \frac{3}{x} - \frac{1}{x^2} \right)} = |x| \sqrt{2 + \frac{3}{x} - \frac{1}{x^2}}$ Again, as $x \to \infty$, we drop the absolute value.

2. Taking the limit as $x \to \infty$: As $x \to \infty$, the terms $\frac{5}{x}$, $\frac{7}{x^2}$, $\frac{3}{x}$, and $\frac{1}{x^2}$ all approach 0. So, the expressions simplify to:

   • Numerator: $\sqrt{8 + 0 + 0} = \sqrt{8}$
   • Denominator: $\sqrt{2 + 0 + 0} = \sqrt{2}$

3. Final limit expression: The limit becomes: $\lim_{{x \to \infty}} \frac{\sqrt{8}}{\sqrt{2}} = \frac{\sqrt{8}}{\sqrt{2}} = \frac{2\sqrt{2}}{\sqrt{2}} = 2$

Thus, the value of the limit is $\boxed{2}$.

Would you like more details or have any further questions?

Here are 5 related questions for deeper understanding:

1. What happens to the limit if the degree of the polynomial in the numerator is higher than in the denominator?
2. How do you approach limits where the terms inside the square roots grow faster than quadratic terms?
3. Can you find the limit if the expression involves cube roots instead of square roots?
4. What if we were finding the limit as $x \to -\infty$ instead?
5. How would the limit change if one of the coefficients inside the square root was negative?

Tip: When evaluating limits of square roots of polynomials as $x \to \infty$, always factor out the highest power of $x$ from under the square root to simplify.