We need to compute the value of the limit:

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

This limit can be solved using the derivative definition of a function. Notice that this expression resembles the definition of the derivative for a function at a specific point. Let's rewrite this limit in terms of derivatives.

Step 1: Recognize the form

Let $f(x) = x^{\frac{1}{4}}$. The expression:

$\frac{(16 + h)^{\frac{1}{4}} - 2}{h}$

is similar to the definition of the derivative of $f(x)$ at $x = 16$:

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

Here, $f(x) = x^{\frac{1}{4}}$, so:

$f(16) = 16^{\frac{1}{4}} = 2$

Therefore, the given limit is just the derivative of $f(x) = x^{\frac{1}{4}}$ evaluated at $x = 16$.

Step 2: Compute the derivative

Now, let's compute the derivative of $f(x) = x^{\frac{1}{4}}$:

$f'(x) = \frac{d}{dx}\left(x^{\frac{1}{4}}\right) = \frac{1}{4} x^{-\frac{3}{4}}$

Evaluating this derivative at $x = 16$:

$f'(16) = \frac{1}{4} \cdot 16^{-\frac{3}{4}} = \frac{1}{4} \cdot \frac{1}{16^{\frac{3}{4}}} = \frac{1}{4} \cdot \frac{1}{8} = \frac{1}{32}$

Step 3: Conclusion

Thus, the value of the limit is:

$\boxed{\frac{1}{32}}$

Would you like more details or have any questions?

Here are five related questions for further practice:

1. What is the derivative of $f(x) = x^{\frac{1}{2}}$ at $x = 25$?
2. How would you evaluate $\lim_{h\to0} \frac{(25+h)^{\frac{1}{2}}-5}{h}$?
3. What is the general form of the derivative for $f(x) = x^{n}$, where $n$ is any rational number?
4. What is the geometric interpretation of the derivative of a function?
5. How would you apply the chain rule to compute the derivative of $f(x) = (3x+2)^{\frac{1}{4}}$?

Tip: Derivatives are powerful for calculating rates of change, especially for functions involving powers and roots!