The limit provided is:

$\lim_{h \to 0} \frac{\sqrt{25 + h} - 5}{h}$

This resembles the definition of the derivative of a function $f(x)$ at some point $a$, which is:

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

Step 1: Recognizing the function $f(x)$

Looking at the given limit expression, it can be compared to the standard derivative definition:

• The expression under the square root, $\sqrt{25 + h}$, suggests that $f(x) = \sqrt{x}$.
• Additionally, we observe that $f(a) = 5$, so $\sqrt{a} = 5$. Solving this gives $a = 25$.

Step 2: Verifying the function

Now that we know $a = 25$, we can set up $f(x) = \sqrt{x}$. Taking the derivative of $f(x)$ at $a = 25$, we can verify the limit:

$f'(x) = \frac{1}{2\sqrt{x}}$ $f'(25) = \frac{1}{2\sqrt{25}} = \frac{1}{2 \times 5} = \frac{1}{10}$

This derivative corresponds to the given limit, confirming that $f(x) = \sqrt{x}$ and $a = 25$.

Final Answer:

• $f(x) = \sqrt{x}$
• $a = 25$

Would you like further details or have any questions? Here are 5 related questions to explore:

1. How do you apply the definition of a derivative to general functions?
2. What are some other examples of limits that represent derivatives?
3. How can you compute limits involving radicals like this one?
4. How do we prove the chain rule using limit definitions of derivatives?
5. What is the geometric meaning of the derivative at a point?

Tip: When handling limits involving square roots, rationalizing the numerator is a helpful technique to simplify the expression.