Discuss the continuity of the function
`f(x) ={:{((Sinx)/(x) ", "x lt0 ),(x+1 ", " x ge0):},`
at =0.

Answer

To discuss the continuity of the function

\[ f(x) = \begin{cases} \frac{\sin x}{x} & \text{if } x < 0 \\ x + 1 & \text{if } x \geq 0 \end{cases} \]

at \( x = 0 \), we need to check the following three conditions:

1. The left-hand limit of \( f(x) \) as \( x \) approaches 0.
2. The right-hand limit of \( f(x) \) as \( x \) approaches 0.
3. The value of \( f(0) \).

### Step 1: Calculate the left-hand limit as \( x \) approaches 0

The left-hand limit is given by:

\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{\sin x}{x} \]

Using the known limit:

\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \]

Thus,

\[ \lim_{x \to 0^-} f(x) = 1 \]

### Step 2: 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 \( x = 0 \):

\[ \lim_{x \to 0^+} f(x) = 0 + 1 = 1 \]

### Step 3: Calculate the value of \( f(0) \)

Since \( f(x) \) is defined as \( x + 1 \) for \( x \geq 0 \):

\[ f(0) = 0 + 1 = 1 \]

### Conclusion

Now we have:

- Left-hand limit: \( \lim_{x \to 0^-} f(x) = 1 \)
- Right-hand limit: \( \lim_{x \to 0^+} f(x) = 1 \)
- Value at \( f(0) = 1 \)

Since the left-hand limit, right-hand limit, and the value of the function at 0 are all equal, we conclude that:

\[ f(x) \text{ is continuous at } x = 0. \]

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