Examine the function for continuity at x = 0 :
`f(x)={{:(sinx/x" when "xlt0),(x+1" when "xge0):}`.

Answer

To examine the function \( f(x) \) for continuity at \( x = 0 \), we will follow these steps:

1. **Identify the function**:
The function is defined as:
\[ f(x) = \begin{cases} \frac{\sin x}{x} & \text{when } x < 0 \\ x + 1 & \text{when } x \geq 0 \end{cases} \]

2. **Find \( f(0) \)**:
Since \( 0 \) falls into the case where \( x \geq 0 \), we use the second part of the function:
\[ f(0) = 0 + 1 = 1 \]

3. **Calculate the left-hand limit as \( x \) approaches 0**:
The left-hand limit is given by:
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{\sin x}{x} \]
We know from the standard limit that:
\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \]
Therefore,
\[ \lim_{x \to 0^-} f(x) = 1 \]

4. **Calculate the right-hand limit as \( x \) approaches 0**:
The right-hand limit is given by:
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x + 1) \]
Substituting \( x = 0 \):
\[ \lim_{x \to 0^+} f(x) = 0 + 1 = 1 \]

5. **Check continuity**:
For the function \( f(x) \) to be continuous at \( x = 0 \), the following must hold:
\[ f(0) = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) \]
We have:
\[ f(0) = 1, \quad \lim_{x \to 0^-} f(x) = 1, \quad \lim_{x \to 0^+} f(x) = 1 \]
Since all three values are equal, we conclude that:
\[ f(x) \text{ is continuous at } x = 0. \]

### Summary of Steps:
1. Identify the function and its parts.
2. Calculate \( f(0) \).
3. Find the left-hand limit as \( x \) approaches 0.
4. Find the right-hand limit as \( x \) approaches 0.
5. Check if all values are equal to conclude continuity.