The problem you shared asks for the value of the following limit:

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

This expression is a derivative calculation in disguise. Let's break it down:

Step-by-Step Explanation:

1. Recognize the form: This limit is the definition of the derivative of the function $f(x) = \sqrt{x}$ at $x = 16$. The formula for a derivative at a point $a$ is: $f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$ Here, $f(x) = \sqrt{x}$, so $f(16) = \sqrt{16} = 4$.

2. Apply the definition of the derivative: $\lim_{h \to 0} \frac{\sqrt{16+h} - 4}{h}$ This matches the form we are given in the problem, except the $\sqrt{16+h} - 4$ part is written as $(16 + h)^{\frac{1}{2}} - 2$ in the problem (since $\sqrt{x} = x^{1/2}$ and $\sqrt{16} = 4$).

3. Evaluate the limit: To solve the limit, we can rationalize the numerator by multiplying the numerator and denominator by the conjugate of the numerator: $\lim_{h \to 0} \frac{(16+h)^{\frac{1}{2}} - 4}{h} \cdot \frac{(16+h)^{\frac{1}{2}} + 4}{(16+h)^{\frac{1}{2}} + 4}$ Simplifying the numerator using the difference of squares: $\lim_{h \to 0} \frac{(16+h) - 16}{h((16+h)^{\frac{1}{2}} + 4)} = \lim_{h \to 0} \frac{h}{h((16+h)^{\frac{1}{2}} + 4)}$ The $h$'s cancel out: $\lim_{h \to 0} \frac{1}{(16+h)^{\frac{1}{2}} + 4}$ Now, substitute $h = 0$: $\frac{1}{\sqrt{16} + 4} = \frac{1}{4 + 4} = \frac{1}{8}$

Final Answer:

$\lim_{h \to 0} \frac{(16+h)^{\frac{1}{2}} - 4}{h} = \frac{1}{8}$

Would you like further clarification or details? Here are some related questions to explore:

1. What is the general form of a derivative using limits?
2. How can rationalizing the numerator help in evaluating limits?
3. Why do we cancel out $h$ in the fraction after rationalizing?
4. What is the significance of evaluating limits when $h \to 0$?
5. How would this process change if the base value $x = 16$ were different?

Tip: Always check if a limit corresponds to the derivative definition. This helps simplify evaluation!