Determine the values of `a , b , c` for which the function `f(x)={(sin(a+1)x+sinx)/(x )` for `x<0`, `f(x)=c` for `x=0`, `f(x)`= ` (sqrt(x+b x^2)-sqrt(x))/(b x^(3/2))`,for `x >0` is continuous at `x=0`

Explanation

To determine the values of \( a, b, c \) for which the function

\[ f(x) = \begin{cases} \frac{\sin(a+1)x + \sin x}{x} & \text{for } x < 0 \\ c & \text{for } x = 0 \\ \frac{\sqrt{x + bx^2} - \sqrt{x}}{b x^{3/2}} & \text{for } x > 0 \end{cases} \]

is continuous at \( x = 0 \), we need to ensure that the left-hand limit \( f(0^-) \), the value at \( x = 0 \) \( f(0) \), and the right-hand limit \( f(0^+) \) are all equal.

### Step 1: Calculate \( f(0^-) \)

We calculate the limit as \( x \) approaches 0 from the left:

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

Using the limit properties of sine, we can split this into two parts:

\[ = \lim_{x \to 0^-} \left( \frac{\sin(a+1)x}{x} + \frac{\sin x}{x} \right) \]

As \( x \to 0 \), \( \frac{\sin x}{x} \to 1 \) and \( \frac{\sin(a+1)x}{x} \to a + 1 \):

\[ = (a + 1) + 1 = a + 2 \]

### Step 2: Calculate \( f(0^+) \)

Now we calculate the limit as \( x \) approaches 0 from the right:

\[ f(0^+) = \lim_{x \to 0^+} \frac{\sqrt{x + bx^2} - \sqrt{x}}{b x^{3/2}} \]

This expression is of the form \( \frac{0}{0} \) as \( x \to 0 \), so we can apply L'Hôpital's Rule or simplify it. We will simplify it by multiplying the numerator and denominator by the conjugate:

\[ = \lim_{x \to 0^+} \frac{(\sqrt{x + bx^2} - \sqrt{x})(\sqrt{x + bx^2} + \sqrt{x})}{b x^{3/2}(\sqrt{x + bx^2} + \sqrt{x})} \]

This simplifies to:

\[ = \lim_{x \to 0^+} \frac{(x + bx^2 - x)}{b x^{3/2}(\sqrt{x + bx^2} + \sqrt{x})} = \lim_{x \to 0^+} \frac{bx^2}{b x^{3/2}(\sqrt{x + bx^2} + \sqrt{x})} \]

Cancelling \( b \) (assuming \( b \neq 0 \)):

\[ = \lim_{x \to 0^+} \frac{x^{1/2}}{\sqrt{x + bx^2} + \sqrt{x}} = \lim_{x \to 0^+} \frac{x^{1/2}}{\sqrt{x(1 + b x)} + \sqrt{x}} = \lim_{x \to 0^+} \frac{x^{1/2}}{\sqrt{x}(1 + \sqrt{b x})} = \lim_{x \to 0^+} \frac{1}{1 + \sqrt{b x}} = \frac{1}{1} = 1 \]

### Step 3: Set the limits equal to each other

For continuity at \( x = 0 \):

\[ f(0^-) = f(0) = f(0^+) \]

From our calculations:

1. \( f(0^-) = a + 2 \)
2. \( f(0) = c \)
3. \( f(0^+) = 1 \)

Setting these equal gives us the equations:

1. \( a + 2 = c \) (Equation 1)
2. \( c = 1 \) (Equation 2)

### Step 4: Solve the equations

From Equation 2, substituting \( c = 1 \) into Equation 1:

\[ a + 2 = 1 \implies a = 1 - 2 = -1 \]

### Step 5: Determine \( b \)

The value of \( b \) can be any real number except \( 0 \) (to avoid division by zero in the expression for \( f(0^+) \)). Therefore, we conclude:

- \( a = -1 \)
- \( b \in \mathbb{R} \setminus \{0\} \)
- \( c = 1 \)

### Final Answer

\[ a = -1, \quad b \in \mathbb{R} \setminus \{0\}, \quad c = 1 \]