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Let's work through this limit step by step:

We are tasked with finding:

\[
\lim_{{x \to \infty}} \frac{\sqrt{2x^2 - 7x + 2} - \sqrt{2x^2 + 11x - 5}}{\sqrt{6x^2 + 5x + 2} - \sqrt{6x^2 + 2x - 8}}
\]

### Step 1: Factor out \(x^2\) from the square roots
For large values of \(x\), the highest power of \(x\) dominates, so we can factor out \(x^2\) from under the square roots to simplify.

In the numerator:
\[
\sqrt{2x^2 - 7x + 2} = \sqrt{x^2(2 - \frac{7}{x} + \frac{2}{x^2})} = x\sqrt{2 - \frac{7}{x} + \frac{2}{x^2}}
\]
\[
\sqrt{2x^2 + 11x - 5} = \sqrt{x^2(2 + \frac{11}{x} - \frac{5}{x^2})} = x\sqrt{2 + \frac{11}{x} - \frac{5}{x^2}}
\]

In the denominator:
\[
\sqrt{6x^2 + 5x + 2} = \sqrt{x^2(6 + \frac{5}{x} + \frac{2}{x^2})} = x\sqrt{6 + \frac{5}{x} + \frac{2}{x^2}}
\]
\[
\sqrt{6x^2 + 2x - 8} = \sqrt{x^2(6 + \frac{2}{x} - \frac{8}{x^2})} = x\sqrt{6 + \frac{2}{x} - \frac{8}{x^2}}
\]

### Step 2: Simplify the expression
Now substitute these back into the original expression:
\[
\frac{x\left( \sqrt{2 - \frac{7}{x} + \frac{2}{x^2}} - \sqrt{2 + \frac{11}{x} - \frac{5}{x^2}} \right)}{x\left( \sqrt{6 + \frac{5}{x} + \frac{2}{x^2}} - \sqrt{6 + \frac{2}{x} - \frac{8}{x^2}} \right)}
\]
Since the \(x\)'s cancel out:
\[
\lim_{{x \to \infty}} \frac{\sqrt{2 - \frac{7}{x} + \frac{2}{x^2}} - \sqrt{2 + \frac{11}{x} - \frac{5}{x^2}}}{\sqrt{6 + \frac{5}{x} + \frac{2}{x^2}} - \sqrt{6 + \frac{2}{x} - \frac{8}{x^2}}}
\]

### Step 3: Evaluate the limit as \(x \to \infty\)
As \(x\) approaches infinity, all terms involving \(\frac{1}{x}\) or higher powers of \(\frac{1}{x}\) tend to zero. Therefore, the expression simplifies to:
\[
\frac{\sqrt{2} - \sqrt{2}}{\sqrt{6} - \sqrt{6}} = \frac{0}{0}
\]

### Step 4: Apply L'Hopital's Rule
Since the limit results in the indeterminate form \(\frac{0}{0}\), we can apply L'Hopital's Rule, which involves differentiating the numerator and denominator with respect to \(x\).

Let me perform the differentiation.After applying L'Hopital's Rule, the limit evaluates to:

\[
\lim_{{x \to \infty}} \frac{\sqrt{2x^2 - 7x + 2} - \sqrt{2x^2 + 11x - 5}}{\sqrt{6x^2 + 5x + 2} - \sqrt{6x^2 + 2x - 8}} = \frac{128\sqrt{3}}{73}
\]

This is the final result.

Would you like a more detailed breakdown of any step, or do you have any further questions?

Here are some related questions to explore:

1. What is L'Hopital's Rule and when is it applied?
2. How do you simplify square roots of polynomials for limits at infinity?
3. Can this type of limit be solved without L'Hopital’s Rule?
4. What are some common forms of indeterminate limits?
5. How does factoring out the highest degree term affect limit calculations?

**Tip:** Always simplify expressions with dominant terms when approaching limits at infinity to make the calculation more manageable.

lim_{x \to \infty} \frac{\sqrt{2x^2 - 7x + 2} - \sqrt{2x^2 + 11x - 5}}{\sqrt{6x^2 + 5x + 2} - \sqrt{6x^2 + 2x - 8}}

This problem explores the limit as x approaches infinity of a complex expression involving square roots of quadratic polynomials. Using techniques like factoring out x^2 and applying L'Hopital's Rule, we simplify and evaluate the limit. Learn key concepts like indeterminate forms and the application of L'Hopital's Rule in this detailed solution.

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