The problem asks to evaluate the following limit as $x \to -\infty$:

$\lim_{x \to -\infty} \left( \sqrt{x^2 + 3x + 1} + x \right)$

Solution:

To handle this limit, we will use the strategy of factoring out the dominant term (which is $x^2$ inside the square root) and simplifying the expression.

1. Simplify the square root expression: Inside the square root, the highest-degree term is $x^2$. So, factor out $x^2$ from the terms inside the square root:

   $\sqrt{x^2 + 3x + 1} = \sqrt{x^2\left(1 + \frac{3}{x} + \frac{1}{x^2}\right)} = |x| \sqrt{1 + \frac{3}{x} + \frac{1}{x^2}}$

   For $x \to -\infty$, $|x| = -x$ (since $x$ is negative). So,

   $\sqrt{x^2 + 3x + 1} = -x \sqrt{1 + \frac{3}{x} + \frac{1}{x^2}}$

2. Rewrite the limit:

   Substitute the simplified square root expression into the original limit:

   $\lim_{x \to -\infty} \left( -x \sqrt{1 + \frac{3}{x} + \frac{1}{x^2}} + x \right)$

   Factor out $x$:

   $\lim_{x \to -\infty} x \left( -\sqrt{1 + \frac{3}{x} + \frac{1}{x^2}} + 1 \right)$

3. Evaluate the limit as $x \to -\infty$:

   As $x \to -\infty$, both $\frac{3}{x}$ and $\frac{1}{x^2}$ approach 0. Therefore,

   $\sqrt{1 + \frac{3}{x} + \frac{1}{x^2}} \to 1$

   This simplifies the expression to:

   $\lim_{x \to -\infty} x \left( -1 + 1 \right) = \lim_{x \to -\infty} x \cdot 0 = 0$

Final Answer:

$\lim_{x \to -\infty} \left( \sqrt{x^2 + 3x + 1} + x \right) = 0$

Let me know if you need further details or clarification.

Related Questions:

1. How do you evaluate limits involving square roots and polynomials as $x \to \infty$?
2. Why does $|x| = -x$ when $x \to -\infty$?
3. What other techniques can simplify limits involving square roots?
4. How can we verify the result of this limit graphically or numerically?
5. What happens if we evaluate this limit as $x \to +\infty$?

Tip:

Always factor out the highest degree of $x$ in complex limits to simplify expressions before taking the limit.