We want to evaluate:

$\lim_{x \to \frac{1}{3}} \left(\sqrt[3]{\frac{x}{9}} - \frac{1}{3}\right).$

Step 1: Simplify the expression

Let $f(x) = \sqrt[3]{\frac{x}{9}} - \frac{1}{3}$. We are tasked to compute the limit as $x \to \frac{1}{3}$.

The cube root can make direct substitution tricky, so we examine the structure of $f(x)$ by simplifying it further.

Step 2: Use substitution for cube root

Let $u = \frac{x}{9}$. Then, $x \to \frac{1}{3}$ corresponds to $u \to \frac{\frac{1}{3}}{9} = \frac{1}{27}$. The function becomes:

$f(x) = \sqrt[3]{u} - \frac{1}{3}.$

Step 3: Expand using Taylor series (or Binomial Approximation)

Near $u = \frac{1}{27}$, the cube root $\sqrt[3]{u}$ can be expanded as:

[ \sqrt[3]{u} = \sqrt[3]{\frac{1}{27}} + \frac{1}{3 \cdot \sqrt[3]{\frac{1}{27}^2}} \cdot (u - \frac{1}{27}) + \text{higher-order terms}. ]

Since $\sqrt[3]{\frac{1}{27}} = \frac{1}{3}$, we rewrite:

$\sqrt[3]{u} - \frac{1}{3} \approx \frac{1}{3 \cdot (\frac{1}{3})^2} (u - \frac{1}{27}) = 9(u - \frac{1}{27}).$

Step 4: Back-substitute $u = \frac{x}{9}$

Replace $u$ with $\frac{x}{9}$, and expand the term $u - \frac{1}{27}$:

$u - \frac{1}{27} = \frac{x}{9} - \frac{1}{27} = \frac{3x - 1}{27}.$

Thus, the function simplifies to:

$f(x) \approx 9 \cdot \frac{3x - 1}{27} = \frac{3x - 1}{3}.$

Step 5: Compute the limit as $x \to \frac{1}{3}$

Substitute $x = \frac{1}{3}$:

$f(x) \to \frac{3 \cdot \frac{1}{3} - 1}{3} = \frac{1 - 1}{3} = 0.$

Final Answer:

$\lim_{x \to \frac{1}{3}} \left(\sqrt[3]{\frac{x}{9}} - \frac{1}{3}\right) = 0.$

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Would you like a detailed explanation of any specific step?

Follow-Up Questions:

1. Why is Taylor expansion useful for evaluating this limit?
2. How can the binomial theorem simplify cube root approximations?
3. What happens if we use L'Hôpital's Rule to evaluate the limit?
4. How does substituting $u = \frac{x}{9}$ simplify the problem?
5. Can you verify this limit numerically for values close to $x = \frac{1}{3}$?

Tip:

When cube roots or roots of higher degrees appear, approximations like Taylor series can simplify complex limits.