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Limite lorsque x tend vers moins l'infinie 

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:

   \[
   \sqrt{x^2 + x + 1} = \sqrt{x^2(1 + \frac{1}{x} + \frac{1}{x^2})}
   = |x| \sqrt{1 + \frac{1}{x} + \frac{1}{x^2}}.
   \]

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

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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.

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

Learn how to evaluate the limit of the expression sqrt(x^2 + x + 1) divided by (x - 2) as x approaches negative infinity. This problem involves asymptotic behavior and square root simplifications, making it ideal for students in advanced high school grades.

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