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

1. Problem Description

We need to find the limit of the function $x + \sqrt{x^2 + 9}$ as $x$ approaches negative infinity.

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

2. Solution Steps

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

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

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

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

We can factor out $x$ from the square root. Since $x \to -\infty$, $\sqrt{x^2} = |x| = -x$.

$\sqrt{x^2 + 9} = \sqrt{x^2(1 + \frac{9}{x^2})} = |x|\sqrt{1 + \frac{9}{x^2}} = -x\sqrt{1 + \frac{9}{x^2}}$

Substitute this back into the limit:

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

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

As $x \to -\infty$, $\frac{9}{x^2} \to 0$. Therefore, $\sqrt{1 + \frac{9}{x^2}} \to \sqrt{1 + 0} = 1$.

So we have:

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

As $x$ approaches negative infinity, $\frac{-9}{2x}$ approaches

0. $\lim_{x\to -\infty} \frac{-9}{2x} = 0$

3. Final Answer

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