To solve the limit:

[ \lim_{x \to \infty} \frac{4 \sqrt[3]{x} - 2\sqrt{x - 1}}{2 \sqrt[3]{x} - 9\sqrt{x + 1}} ]

we proceed step by step:

Step 1: Simplify the dominant terms

As $x \to \infty$, the dominant term in each part of the numerator and denominator is the one with the largest growth rate. Here:

• $\sqrt[3]{x}$ grows slower than $\sqrt{x}$, so $\sqrt{x}$ will dominate as $x \to \infty$.

Step 2: Factor out the dominant term

In both numerator and denominator, factor out $\sqrt{x}$ from terms involving square roots. Rewrite:

$\sqrt{x - 1} = \sqrt{x(1 - \frac{1}{x})} = \sqrt{x} \cdot \sqrt{1 - \frac{1}{x}}$

and similarly for $\sqrt{x + 1}$. The expression becomes:

[ \frac{4\sqrt[3]{x} - 2\sqrt{x}(1 - \frac{1}{x})^{1/2}}{2\sqrt[3]{x} - 9\sqrt{x}(1 + \frac{1}{x})^{1/2}} ]

Step 3: Simplify the ratio

As $x \to \infty$, the square root terms ($\sqrt{1 - \frac{1}{x}}$ and $\sqrt{1 + \frac{1}{x}}$) approach 1. After substitution:

[ \frac{4\sqrt[3]{x} - 2\sqrt{x}}{2\sqrt[3]{x} - 9\sqrt{x}} ]

Factor $\sqrt{x}$ out of the numerator and denominator:

[ \frac{\sqrt{x} \left( \frac{4}{\sqrt[3]{x}} - 2 \right)}{\sqrt{x} \left( \frac{2}{\sqrt[3]{x}} - 9 \right)} ]

Step 4: Evaluate the limit

As $x \to \infty$, $\frac{1}{\sqrt[3]{x}} \to 0$. The remaining terms are:

$\frac{-2}{-9} = \frac{2}{9}$

Thus, the limit is:

$\boxed{\frac{2}{9}}$

Would you like me to elaborate on any step or confirm understanding?

Relative Questions:

1. What is the difference between growth rates of $\sqrt{x}$ and $\sqrt[3]{x}$?
2. Why does $\frac{1}{x}$ vanish as $x \to \infty$?
3. How do dominant terms help simplify complex expressions?
4. What are other methods to evaluate limits involving square roots and cube roots?
5. Can this process be applied to limits approaching zero?

Tip:

Always factor out the dominant term when evaluating limits at infinity to simplify expressions and identify key contributions.