We need to evaluate the limit of the expression $(x + \sqrt{x^2 - 9})$ as $x$ approaches negative infinity. $$c = \lim_{x \to -\infty} (x + \sqrt{x^2 - 9})$$

1. Problem Description

We need to evaluate the limit of the expression $(x + \sqrt{x^2 - 9})$ as $x$ approaches negative infinity.

$$c = \lim_{x \to -\infty} (x + \sqrt{x^2 - 9})$$

2. Solution Steps

To evaluate this limit, we can multiply and divide by the conjugate of the expression $x + \sqrt{x^2 - 9}$.

$$c = \lim_{x \to -\infty} (x + \sqrt{x^2 - 9}) \cdot \frac{x - \sqrt{x^2 - 9}}{x - \sqrt{x^2 - 9}}$$

$$c = \lim_{x \to -\infty} \frac{x^2 - (x^2 - 9)}{x - \sqrt{x^2 - 9}}$$

$$c = \lim_{x \to -\infty} \frac{9}{x - \sqrt{x^2 - 9}}$$

Now, let's factor out $x^2$ from the square root. Since $x \to -\infty$, we have $\sqrt{x^2} = |x| = -x$.

$$c = \lim_{x \to -\infty} \frac{9}{x - \sqrt{x^2(1 - \frac{9}{x^2})}}$$

$$c = \lim_{x \to -\infty} \frac{9}{x - |x|\sqrt{1 - \frac{9}{x^2}}}$$

Since $x \to -\infty$, $|x| = -x$, so

$$c = \lim_{x \to -\infty} \frac{9}{x - (-x)\sqrt{1 - \frac{9}{x^2}}}$$

$$c = \lim_{x \to -\infty} \frac{9}{x + x\sqrt{1 - \frac{9}{x^2}}}$$

$$c = \lim_{x \to -\infty} \frac{9}{x(1 + \sqrt{1 - \frac{9}{x^2}})}$$

As $x \to -\infty$, $\frac{9}{x^2} \to 0$, so $\sqrt{1 - \frac{9}{x^2}} \to \sqrt{1} = 1$. Thus we have:

$$c = \lim_{x \to -\infty} \frac{9}{x(1 + 1)} = \lim_{x \to -\infty} \frac{9}{2x}$$

As $x \to -\infty$, the fraction $\frac{9}{2x}$ goes to $0$.

$$c = 0$$

3. Final Answer

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