The problem asks us to find the limit as $x$ approaches infinity of the expression $\frac{2x - \sqrt{2x^2-1}}{4x - 3\sqrt{x^2+2}}$.

1. Problem Description

The problem asks us to find the limit as $x$ approaches infinity of the expression $\frac{2x - \sqrt{2x^2-1}}{4x - 3\sqrt{x^2+2}}$.

2. Solution Steps

First, we divide both the numerator and the denominator by $x$.

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

Now, we bring $x$ inside the square roots. Since $x>0$ as $x\to \infty$, we have $x = \sqrt{x^2}$.

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

As $x$ approaches infinity, $\frac{1}{x^2}$ approaches 0 and $\frac{2}{x^2}$ approaches

0.

$\lim_{x\to\infty} \frac{2 - \sqrt{2 - \frac{1}{x^2}}}{4 - 3\sqrt{1 + \frac{2}{x^2}}} = \frac{2 - \sqrt{2-0}}{4 - 3\sqrt{1+0}} = \frac{2 - \sqrt{2}}{4 - 3(1)} = \frac{2 - \sqrt{2}}{4 - 3} = \frac{2 - \sqrt{2}}{1} = 2 - \sqrt{2}$

3. Final Answer

$2 - \sqrt{2}$