Evaluate `lim_(x to 0) [(2008 "sin x)/(x)] + [(2009 tan x)/(x)]` where [x] denotes the greatest integer `le x`.

Answer

To evaluate the limit

\[ \lim_{x \to 0} \left[ \frac{2008 \sin x}{x} + \frac{2009 \tan x}{x} \right], \]

we start by using the known limits of \(\frac{\sin x}{x}\) and \(\frac{\tan x}{x}\) as \(x\) approaches 0.

### Step 1: Use the limits of \(\sin x\) and \(\tan x\)

We know that:

\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \quad \text{and} \quad \lim_{x \to 0} \frac{\tan x}{x} = 1. \]

### Step 2: Substitute the limits into the expression

Using these limits, we can substitute into our limit expression:

\[ \lim_{x \to 0} \left[ \frac{2008 \sin x}{x} + \frac{2009 \tan x}{x} \right] = 2008 \cdot \lim_{x \to 0} \frac{\sin x}{x} + 2009 \cdot \lim_{x \to 0} \frac{\tan x}{x}. \]

### Step 3: Calculate the limit

Substituting the known limits:

\[ = 2008 \cdot 1 + 2009 \cdot 1 = 2008 + 2009 = 4017. \]

### Step 4: Consider the greatest integer function

Since the problem states that \(x\) denotes the greatest integer function, we need to apply the greatest integer function to our result.

The greatest integer function \([x]\) gives the largest integer less than or equal to \(x\). Thus, we have:

\[ \lfloor 4017 \rfloor = 4017. \]

### Final Answer

Therefore, the final answer is:

\[ \boxed{4017}. \]