We are given a limit expression that represents the derivative $f'(a)$. To find $f(x)$ and $a$, we can recognize that the limit represents the definition of the derivative:

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

The given limit expression is:

$\lim_{h \to 0} \frac{\sin\left(\frac{5\pi}{3} + h\right) + \frac{3}{\sqrt{2}}}{h}$

This suggests the form of the derivative of some function $f(x)$, where $f(x) = \sin(x) + \frac{3}{\sqrt{2}}x$, evaluated at $a = \frac{5\pi}{3}$. Let's break this down:

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

In the expression, the term inside the sine function matches the argument $\frac{5\pi}{3} + h$, which suggests that $f(x) = \sin(x)$ is part of the function. The other term $\frac{3}{\sqrt{2}}$ appears as a constant multiple of $h$, implying a linear term in the function.

Thus, we can identify the function as:

$f(x) = \sin(x) + \frac{3}{\sqrt{2}} x$

Step 2: Identify $a$

From the sine term $\frac{5\pi}{3} + h$, we recognize that the point $a$ at which the derivative is being evaluated is $a = \frac{5\pi}{3}$.

Final Answer

The function $f(x)$ and point $a$ are:

$f(x) = \sin(x) + \frac{3}{\sqrt{2}} x$ $a = \frac{5\pi}{3}$

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Would you like further details on how to verify this or explore its derivative?

Here are 5 related questions to consider:

1. What is the derivative of $\sin(x) + \frac{3}{\sqrt{2}}x$?
2. How does the chain rule apply when differentiating trigonometric functions?
3. Can you find the second derivative of $f(x)$ at $a$?
4. How would the limit change if $f(x)$ included a quadratic term?
5. What happens if we change the constant $\frac{3}{\sqrt{2}}$?

Tip: The derivative of a sine function $\sin(x)$ is $\cos(x)$, while the derivative of a linear term is its coefficient.