We are asked to find the limit:

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

Step-by-Step Solution:

1. Behavior of the Square Root: As $x \to -\infty$, the dominant term inside the square root is $x^2$, so we can factor out $x^2$ from the expression under the square root:

   = |x| \sqrt{1 + \frac{1}{x} + \frac{1}{x^2}}.$$ Since $$x \to -\infty$$, $$|x| = -x$$, so: $$\sqrt{x^2 + x + 1} \approx -x \sqrt{1 + \frac{1}{x} + \frac{1}{x^2}}.$$ As $$x \to -\infty$$, both $$\frac{1}{x}$$ and $$\frac{1}{x^2}$$ approach zero, so: $$\sqrt{x^2 + x + 1} \approx -x.$$

2. Simplifying the Limit Expression: Now, substitute this approximation into the limit expression:

   $\frac{\sqrt{x^2 + x + 1}}{x - 2} \approx \frac{-x}{x - 2}.$

3. Final Simplification: Divide both the numerator and denominator by $x$:

   $\frac{-x}{x - 2} = \frac{-x}{x(1 - \frac{2}{x})} = \frac{-1}{1 - \frac{2}{x}}.$

   As $x \to -\infty$, $\frac{2}{x} \to 0$, so:

   $\frac{-1}{1 - 0} = -1.$

Final Answer:

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

Would you like further details or clarifications?

Here are some related questions to explore:

1. How would the limit change if $x \to +\infty$?
2. What happens if we replace $x^2$ with a higher degree polynomial?
3. What is the geometric interpretation of this limit?
4. How does the behavior of the function differ near $x = 2$?
5. Can L'Hopital's Rule be applied in this case?

Tip: When dealing with limits involving square roots, always factor out the highest power of $x$ from inside the square root to simplify the expression.