The problem you shared is about finding a function $f(x)$ and a value $a$ based on the given limit:

$\lim_{h \to 0} \frac{\ln(e + h) - 1}{h}$

This limit looks like the definition of the derivative for some function evaluated at a particular point $a$, i.e.,

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

From the expression, we can recognize the form:

$\lim_{h \to 0} \frac{\ln(e + h) - 1}{h}$

We can see that:

• The term $\ln(e + h)$ suggests that the function $f(x) = \ln(x)$, since its derivative is $\frac{1}{x}$.
• The constant 1 in the limit corresponds to $\ln(e)$, which is the value of the function at $x = e$. This means $a = e$.

Thus:

• The function $f(x) = \ln(x)$.
• The value $a = e$.

Would you like further details on how this is derived or have any questions? Here are some related questions for practice:

1. What is the derivative of $f(x) = \ln(x)$?
2. How would you set up the limit definition of the derivative for another function, like $f(x) = \sqrt{x}$?
3. What is the importance of recognizing the form of limits as derivative definitions?
4. How does the natural logarithm function behave around $x = e$?
5. Could you apply the same limit process to find the derivative of $f(x) = \sin(x)$ at some value $a$?

Tip: Always try to recognize the structure of limits that represent derivatives—it’s a useful way to solve many calculus problems quickly!