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

1. Problem Description

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

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

2. Solution Steps

To evaluate this limit, we divide both the numerator and the denominator by the highest power of $x$ that appears in the denominator, which is $x$.

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

Since $x = \sqrt{x^2}$ and $x = \sqrt[3]{x^3}$ when $x>0$, we have:

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

As $x \to \infty$, $\frac{1}{x^2} \to 0$, $\frac{1}{x} \to 0$, and $\frac{2}{x^3} \to 0$.

Therefore, the limit becomes:

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

3. Final Answer

$\frac{1}{4}$