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

Answer

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

1. The function \( f(x) \) must be defined at \( x = 0 \).
2. The left-hand limit \( \lim_{x \to 0^-} f(x) \) must exist.
3. The right-hand limit \( \lim_{x \to 0^+} f(x) \) must exist.
4. The left-hand limit, right-hand limit, and the function value at that point must all be equal.

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

### Step 1: Check if \( f(0) \) is defined
From the definition of the function, we see that:
\[ f(0) = 2 \]
Thus, \( f(0) \) is defined.

### Step 2: Calculate the left-hand limit as \( x \) approaches 0
We need to find:
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{\sin x}{x} \]
Using the well-known limit:
\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \]
Thus,
\[ \lim_{x \to 0^-} f(x) = 1 \]

### Step 3: Calculate the right-hand limit as \( x \) approaches 0
Now we find:
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{\sin x}{x} \]
Again, using the same limit:
\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \]
Thus,
\[ \lim_{x \to 0^+} f(x) = 1 \]

### Step 4: Compare the limits and the function value
Now we compare the left-hand limit, right-hand limit, and the function value at \( x = 0 \):
- Left-hand limit: \( \lim_{x \to 0^-} f(x) = 1 \)
- Right-hand limit: \( \lim_{x \to 0^+} f(x) = 1 \)
- Function value: \( f(0) = 2 \)

Since:
\[ \lim_{x \to 0^-} f(x) \neq f(0) \]
and
\[ \lim_{x \to 0^+} f(x) \neq f(0) \]

### Conclusion
The function \( f(x) \) is not continuous at \( x = 0 \).

---