Let's solve the given limit step by step. The problem is:

[ \lim_{x \to 0} \frac{\sqrt[3]{27 + x} - \sqrt[3]{27 - x}}{\sqrt[3]{x^2} + \sqrt[5]{x} + x}. ]

Step 1: Simplify the numerator

The numerator involves a difference of cube roots: [ \sqrt[3]{27 + x} - \sqrt[3]{27 - x}. ]

Using the linear approximation for cube roots, when $x$ is small: [ \sqrt[3]{27 + x} \approx 3 + \frac{x}{27}, \quad \text{and} \quad \sqrt[3]{27 - x} \approx 3 - \frac{x}{27}. ]

So: [ \sqrt[3]{27 + x} - \sqrt[3]{27 - x} \approx \left(3 + \frac{x}{27}\right) - \left(3 - \frac{x}{27}\right) = \frac{2x}{27}. ]

Step 2: Approximate the denominator

The denominator is: [ \sqrt[3]{x^2} + \sqrt[5]{x} + x. ]

When $x \to 0$, the dominant term is the largest power of $x$. Breaking it down:

1. $\sqrt[3]{x^2} = x^{2/3}$,
2. $\sqrt[5]{x} = x^{1/5}$,
3. $x = x^{1}$.

As $x \to 0$, the term $x^{1/5}$ dominates, so the denominator approximates to: [ \sqrt[3]{x^2} + \sqrt[5]{x} + x \approx x^{1/5}. ]

Step 3: Combine numerator and denominator

Substituting the approximations: [ \frac{\sqrt[3]{27 + x} - \sqrt[3]{27 - x}}{\sqrt[3]{x^2} + \sqrt[5]{x} + x} \approx \frac{\frac{2x}{27}}{x^{1/5}}. ]

Simplify the fraction: $\frac{\frac{2x}{27}}{x^{1/5}} = \frac{2}{27} \cdot x^{1 - 1/5} = \frac{2}{27} \cdot x^{4/5}.$

Step 4: Evaluate the limit as $x \to 0$

Since $x^{4/5} \to 0$ as $x \to 0$, the entire expression approaches 0: [ \lim_{x \to 0} \frac{\sqrt[3]{27 + x} - \sqrt[3]{27 - x}}{\sqrt[3]{x^2} + \sqrt[5]{x} + x} = 0. ]

Final Answer:

$\boxed{0}$

Would you like further clarification or details on any step?

Expanding Learning:

1. Why does $x^{1/5}$ dominate the denominator as $x \to 0$?
2. How is linear approximation for cube roots derived?
3. Could L'Hôpital's Rule be used here? Why or why not?
4. What happens if we expand higher-order terms for the cube root approximation?
5. Can this limit be solved graphically or numerically for confirmation?

Tip:

Always identify the dominant terms in a fraction as $x \to 0$ or $x \to \infty$ to simplify limits effectively.