To solve the limit as $x \to -1$ for the given expression:

$\lim_{x \to -1} \frac{2x\sqrt{5} + 3x\sqrt{4} - 5x\sqrt{3} + x\sqrt{2} + x - 6}{x\sqrt{5} - 2x\sqrt{4} + x\sqrt{3} - 6x\sqrt{2} - x + 9}$

Step-by-step process:

1. Substitute $x = -1$ into the expression:

For the numerator: $2(-1)\sqrt{5} + 3(-1)\sqrt{4} - 5(-1)\sqrt{3} + (-1)\sqrt{2} + (-1) - 6$ Simplifying: $-2\sqrt{5} - 6 + 5\sqrt{3} - \sqrt{2} - 1 - 6$ $= -2\sqrt{5} + 5\sqrt{3} - \sqrt{2} - 13$

For the denominator: $(-1)\sqrt{5} - 2(-1)\sqrt{4} + (-1)\sqrt{3} - 6(-1)\sqrt{2} - (-1) + 9$ Simplifying: $-\sqrt{5} + 4 - \sqrt{3} + 6\sqrt{2} + 1 + 9$ $= -\sqrt{5} - \sqrt{3} + 6\sqrt{2} + 14$

2. Final expression after substitution: $\lim_{x \to -1} \frac{-2\sqrt{5} + 5\sqrt{3} - \sqrt{2} - 13}{-\sqrt{5} - \sqrt{3} + 6\sqrt{2} + 14}$

Since we substituted $x = -1$ directly into the expression and did not encounter any indeterminate forms, this is the value of the limit.

Would you like more details or clarification on any part?

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Here are 5 related questions to expand your understanding:

1. How do you simplify expressions with square roots involving variables?
2. What methods can be used to solve limits involving indeterminate forms?
3. How does direct substitution work when evaluating limits?
4. Can you explain when it's necessary to use L'Hôpital's Rule for limits?
5. What is the significance of limits in calculus, particularly at specific points?

Tip: Always check if direct substitution works first when solving limits before using other methods like factoring or L'Hôpital's Rule.