The problem you're dealing with is the following limit:

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

Step 1: Simplify the expression

We can begin by simplifying the expression inside the square root. We focus on the dominant term as $x \to -\infty$. Let's factor $x^2$ out of the square root:

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

Since $x$ is approaching negative infinity, $|x| = -x$. Therefore:

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

Step 2: Substituting this back into the original expression

Now, substitute this into the original expression:

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

Factor out $x$ from the denominator:

$\frac{1}{x(-\sqrt{1 + \frac{2}{x}} + 1)}$

Step 3: Simplify the denominator as $x \to -\infty$

As $x \to -\infty$, the term $\frac{2}{x}$ approaches 0. Therefore, we have:

$\sqrt{1 + \frac{2}{x}} \to 1$

Thus, the denominator becomes:

$x(1 - 1) = 0$

Step 4: Final evaluation

The denominator tends toward $0$ as $x \to -\infty$, and since the numerator is constantly 1, the whole expression behaves like:

$\frac{1}{0} \to 0$

Therefore, the limit is:

$\lim_{x \to -\infty} \frac{1}{\sqrt{x^2 + 2x} + x} = 0$

Summary

The limit is $0$.

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Would you like further details on the steps or have any questions on specific parts of the solution? Here are some related topics to explore:

1. How to handle limits involving square roots in the denominator?
2. What happens when the expression involves higher powers of $x$?
3. How does factoring affect limit evaluation?
4. Why is the denominator approaching zero and how does it influence the limit?
5. What is the role of dominant terms in limits at infinity?

Tip: When dealing with square roots of quadratic expressions, always consider factoring out the dominant term (like $x^2$) to simplify the limit process.