Examine the continuity of the function f(x) at x = 0.
`f(x)={{:((tan2x)/(3x)"when "xne0),(3/2" when "x = 0):}`

Answer

To examine the continuity of the function \( f(x) \) at \( x = 0 \), we need to check the following condition:

A function \( f(x) \) is continuous at a point \( x = a \) if:
1. \( f(a) \) is defined.
2. The left-hand limit \( \lim_{x \to a^-} f(x) \) exists.
3. The right-hand limit \( \lim_{x \to a^+} f(x) \) exists.
4. All three values are equal: \( f(a) = \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) \).

Given the function:
\[ f(x) = \begin{cases} \frac{\tan(2x)}{3x} & \text{when } x \neq 0 \\ \frac{3}{2} & \text{when } x = 0 \end{cases} \]

### Step 1: Evaluate \( f(0) \)
We first find \( f(0) \):
\[ f(0) = \frac{3}{2} \]

### Step 2: Calculate the Left-Hand Limit
Next, we calculate the left-hand limit as \( x \) approaches 0:
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{\tan(2x)}{3x} \]
To evaluate this limit, we can use the fact that \( \tan(2x) \) can be approximated by \( 2x \) as \( x \) approaches 0:
\[ \lim_{x \to 0^-} \frac{\tan(2x)}{3x} = \lim_{x \to 0^-} \frac{2x}{3x} = \lim_{x \to 0^-} \frac{2}{3} = \frac{2}{3} \]

### Step 3: Calculate the Right-Hand Limit
Now, we calculate the right-hand limit as \( x \) approaches 0:
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{\tan(2x)}{3x} \]
Using the same approximation as before:
\[ \lim_{x \to 0^+} \frac{\tan(2x)}{3x} = \lim_{x \to 0^+} \frac{2x}{3x} = \lim_{x \to 0^+} \frac{2}{3} = \frac{2}{3} \]

### Step 4: Compare the Values
Now we compare the values:
- \( f(0) = \frac{3}{2} \)
- \( \lim_{x \to 0^-} f(x) = \frac{2}{3} \)
- \( \lim_{x \to 0^+} f(x) = \frac{2}{3} \)

Since \( f(0) = \frac{3}{2} \) is not equal to \( \frac{2}{3} \), we conclude that:
\[ f(0) \neq \lim_{x \to 0^-} f(x) \quad \text{and} \quad f(0) \neq \lim_{x \to 0^+} f(x) \]

### Conclusion
Thus, the function \( f(x) \) is **not continuous** at \( x = 0 \).