We are tasked with finding the limit:

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

Step-by-step solution:

1. Expression setup: The expression inside the limit is:

   $f(x) = x + \sqrt{x^2 + 8x}$

2. Simplify the square root: The term inside the square root is $x^2 + 8x$. As $x \to -\infty$, the $x^2$ term dominates over $8x$, so we focus on simplifying the square root expression.

   $\sqrt{x^2 + 8x}$

   Factor out $x^2$ from the terms inside the square root to make the expression more manageable:

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

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

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

3. Substitute this back into the original expression: Now substitute this back into the original limit expression:

   $f(x) = x + (-x \sqrt{1 + \frac{8}{x}})$ Simplifying:

   $f(x) = x(1 - \sqrt{1 + \frac{8}{x}})$

4. Evaluate the limit as $x \to -\infty$: Now, consider the behavior of the term $\sqrt{1 + \frac{8}{x}}$ as $x \to -\infty$. Since $\frac{8}{x} \to 0$ as $x \to -\infty$, we have:

   $\sqrt{1 + \frac{8}{x}} \to \sqrt{1} = 1$

   Thus, the expression inside the parentheses becomes:

   $1 - \sqrt{1 + \frac{8}{x}} \to 1 - 1 = 0$

   Therefore:

   $f(x) \to x \times 0 = 0$

Final Answer:

The limit exists and is:

$\boxed{0}$

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Would you like further clarification on any steps or have additional questions? Here are some related questions:

1. How would the result change if $x^2 + 8x$ were under a different root, such as the cube root?
2. What happens if the term inside the square root had a negative coefficient?
3. How does the behavior of the function change if $x \to +\infty$ instead of $x \to -\infty$?
4. Can we apply a similar approach for limits involving more complicated polynomials under a square root?
5. What is the role of the $x^2$ term when taking limits to infinity?

Tip: When simplifying square roots with polynomials, always consider factoring out the highest degree term to make the limit easier to evaluate.