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To solve the given limit problem:

\[
\lim_{{h \to 0}} \frac{{(16 + h)^{\frac{1}{4}} - 2}}{h}
\]

we can use the derivative definition. Let's denote \( f(x) = x^{\frac{1}{4}} \). Then the derivative \( f'(x) \) at \( x = 16 \) is given by:

\[
f'(16) = \lim_{{h \to 0}} \frac{{f(16 + h) - f(16)}}{h}
\]

We know that \( f(16) = 2 \), so we have:

\[
f'(16) = \lim_{{h \to 0}} \frac{{(16 + h)^{\frac{1}{4}} - 2}}{h}
\]

Now, we can find \( f'(x) \) in general. Using the power rule:

\[
f'(x) = \frac{1}{4} x^{-\frac{3}{4}}
\]

At \( x = 16 \):

\[
f'(16) = \frac{1}{4} \times 16^{-\frac{3}{4}}
\]

Calculating \( 16^{-\frac{3}{4}} \):

\[
16 = 2^4 \quad \text{so} \quad 16^{\frac{1}{4}} = 2 \quad \text{and} \quad 16^{-\frac{3}{4}} = 2^{-3} = \frac{1}{8}
\]

Thus:

\[
f'(16) = \frac{1}{4} \times \frac{1}{8} = \frac{1}{32}
\]

The correct answer is:

**Option B: \( \frac{1}{32} \)**

Would you like more details on this solution or any further questions? 

Here are some related questions:
1. How do we find the derivative of a function using the definition of limits?
2. Why does \( (16 + h)^{\frac{1}{4}} \) approximate to 2 when \( h \) is close to 0?
3. How do we compute the derivative of a function of the form \( x^{\frac{1}{n}} \)?
4. What is the geometric interpretation of a derivative at a point?
5. Can the limit definition of the derivative be used for all functions?

**Tip:** When working with limits involving roots, it can be helpful to rationalize the expression to simplify calculations.

What is the value of \( \lim_{{h \to 0}} \frac{{(16 + h)^{\frac{1}{4}} - 2}}{h} \)?

Learn how to evaluate the limit \( \lim_{{h \to 0}} \frac{{(16 + h)^{1/4} - 2}}{h} \) by applying the derivative definition and power rule. This solution is ideal for advanced high school calculus students.

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