We are asked to find the limit of the expression $\frac{2x - \sqrt{2x^2 - 1}}{4x - 3\sqrt{x^2 + 2}}$ as $x$ approaches infinity.

1. Problem Description

We are asked to find the limit of the expression $\frac{2x - \sqrt{2x^2 - 1}}{4x - 3\sqrt{x^2 + 2}}$ as $x$ approaches infinity.

2. Solution Steps

To solve this limit, we can divide both the numerator and denominator by $x$. Note that for $x > 0$, $\sqrt{x^2} = x$.

$\lim_{x \to \infty} \frac{2x - \sqrt{2x^2 - 1}}{4x - 3\sqrt{x^2 + 2}} = \lim_{x \to \infty} \frac{2x - \sqrt{x^2(2 - \frac{1}{x^2})}}{4x - 3\sqrt{x^2(1 + \frac{2}{x^2})}} = \lim_{x \to \infty} \frac{2x - x\sqrt{2 - \frac{1}{x^2}}}{4x - 3x\sqrt{1 + \frac{2}{x^2}}}$

Now, we can factor out $x$ from both the numerator and the denominator:

$= \lim_{x \to \infty} \frac{x(2 - \sqrt{2 - \frac{1}{x^2}})}{x(4 - 3\sqrt{1 + \frac{2}{x^2}})} = \lim_{x \to \infty} \frac{2 - \sqrt{2 - \frac{1}{x^2}}}{4 - 3\sqrt{1 + \frac{2}{x^2}}}$

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

$= \frac{2 - \sqrt{2 - 0}}{4 - 3\sqrt{1 + 0}} = \frac{2 - \sqrt{2}}{4 - 3\sqrt{1}} = \frac{2 - \sqrt{2}}{4 - 3} = \frac{2 - \sqrt{2}}{1} = 2 - \sqrt{2}$

3. Final Answer

$2 - \sqrt{2}$