We are asked to evaluate the limit of the given expression as $x$ approaches infinity: $$ \lim_{x \to \infty} \frac{\sqrt{3x^2+6}}{5-2x} $$

1. Problem Description

We are asked to evaluate the limit of the given expression as $x$ approaches infinity:

$$$$

\lim_{x \to \infty} \frac{\sqrt{3x^2+6}}{5-2x}

$$$$

2. Solution Steps

First, we can divide both the numerator and denominator by $x$. Note that $\sqrt{x^2} = |x|$. Since $x \to \infty$, we have $x>0$, so $|x| = x$. Therefore, we can write $x = \sqrt{x^2}$. Thus, we can rewrite the limit as follows:

$$$$

\lim_{x \to \infty} \frac{\sqrt{3x^2+6}}{5-2x} = \lim_{x \to \infty} \frac{\frac{\sqrt{3x^2+6}}{x}}{\frac{5-2x}{x}} = \lim_{x \to \infty} \frac{\frac{\sqrt{3x^2+6}}{\sqrt{x^2}}}{\frac{5}{x}-\frac{2x}{x}}

$$$$

$$$$

= \lim_{x \to \infty} \frac{\sqrt{\frac{3x^2+6}{x^2}}}{\frac{5}{x}-2} = \lim_{x \to \infty} \frac{\sqrt{3+\frac{6}{x^2}}}{\frac{5}{x}-2}

$$$$

As $x \to \infty$, $\frac{6}{x^2} \to 0$ and $\frac{5}{x} \to 0$. Therefore, we have

$$$$

\lim_{x \to \infty} \frac{\sqrt{3+\frac{6}{x^2}}}{\frac{5}{x}-2} = \frac{\sqrt{3+0}}{0-2} = \frac{\sqrt{3}}{-2} = -\frac{\sqrt{3}}{2}

$$$$

3. Final Answer

The final answer is $-\frac{\sqrt{3}}{2}$