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

1. Problem Description

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

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

2. Solution Steps

We can divide both the numerator and the denominator by $x$. Also, recall that $\sqrt{x^2} = |x|$, and since $x \to \infty$, $x$ is positive, so $|x| = x$. Therefore, we can write $x = \sqrt{x^2}$ when taking it into the square root.

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

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

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

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

So,

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

3. Final Answer

$2-\sqrt{2}$