We are asked to find the following limit:

$\lim_{x \to -\infty} \left( x + \sqrt{x^2 + 8x} \right)$

Step 1: Analyze the expression

As $x$ approaches negative infinity, both $x$ and $\sqrt{x^2 + 8x}$ will behave in specific ways. Let’s break it down to make the limit easier to evaluate.

Step 2: Simplify the square root term

The square root term is $\sqrt{x^2 + 8x}$. We can factor out $x^2$ from inside the square root to better understand its behavior for large negative $x$:

$\sqrt{x^2 + 8x} = \sqrt{x^2(1 + \frac{8}{x})} = |x| \sqrt{1 + \frac{8}{x}}$

For $x \to -\infty$, we know that $|x| = -x$ because $x$ is negative. Therefore:

$\sqrt{x^2 + 8x} = -x \sqrt{1 + \frac{8}{x}}$

Step 3: Approximate for large negative $x$

As $x \to -\infty$, the term $\frac{8}{x}$ becomes very small. Therefore, we can use the approximation for $\sqrt{1 + \frac{8}{x}}$:

$\sqrt{1 + \frac{8}{x}} \approx 1 + \frac{4}{x}$

Thus, we have:

$\sqrt{x^2 + 8x} \approx -x \left( 1 + \frac{4}{x} \right) = -x - 4$

Step 4: Substitute into the original expression

Now substitute this approximation into the original expression:

$x + \sqrt{x^2 + 8x} \approx x + (-x - 4) = -4$

Step 5: Conclude the limit

Thus, the limit is:

$\lim_{x \to -\infty} \left( x + \sqrt{x^2 + 8x} \right) = -4$

Final Answer:

The limit exists and is $\boxed{-4}$.

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Would you like further explanation or have any additional questions?

Here are 5 related questions for you:

1. How does factoring out $x^2$ help in evaluating limits involving square roots?
2. What happens if we have a different expression inside the square root, such as $x^2 + 3x$?
3. Can this method be applied to limits where $x$ approaches positive infinity?
4. What is the behavior of $\sqrt{x^2 + 8x}$ as $x \to \infty$?
5. How can we determine the limit of a rational function as $x \to -\infty$?

Tip: When working with square roots of quadratic expressions, it's helpful to factor out $x^2$ to simplify the expression and identify leading terms for large values of $x$.