If `f(x) ={{:((sin(a+1)x + sinx)/x, "for " x lt 0),(=c, "for " x=0),(((x+bx^(2))^(1//2)-x^(1//2))/(bx^(1//2)), "for " x gt 0):}` is continous at x=0, then a=……………, b=………. , c=………..

Answer

To determine the values of \( a \), \( b \), and \( c \) such that the function \( f(x) \) is continuous at \( x = 0 \), we need to analyze the left-hand limit and the right-hand limit of \( f(x) \) as \( x \) approaches \( 0 \).

### Step 1: Define the Function
The function is defined as follows:
\[ f(x) = \begin{cases} \frac{\sin(a+1)x + \sin x}{x} & \text{for } x < 0 \\ c & \text{for } x = 0 \\ \frac{(x + bx^2)^{1/2} - x^{1/2}}{b x^{1/2}} & \text{for } x > 0 \end{cases} \]

### Step 2: Calculate the Left-Hand Limit as \( x \to 0^- \)
We need to find:
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{\sin(a+1)x + \sin x}{x} \]
Using the limit properties, we can rewrite this as:
\[ \lim_{x \to 0^-} \left( \frac{\sin(a+1)x}{x} + \frac{\sin x}{x} \right) \]
Using the fact that \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \), we have:
\[ \lim_{x \to 0^-} \frac{\sin(a+1)x}{x} = a + 1 \]
Thus, the left-hand limit becomes:
\[ \lim_{x \to 0^-} f(x) = (a + 1) + 1 = a + 2 \]

### Step 3: Calculate the Right-Hand Limit as \( x \to 0^+ \)
Now we find:
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{(x + bx^2)^{1/2} - x^{1/2}}{b x^{1/2}} \]
To simplify this, we can factor out \( x^{1/2} \):
\[ = \lim_{x \to 0^+} \frac{x^{1/2} \left( (1 + b x)^{1/2} - 1 \right)}{b x^{1/2}} = \frac{1}{b} \lim_{x \to 0^+} \left( (1 + b x)^{1/2} - 1 \right) \]
Using the binomial expansion for small \( x \):
\[ (1 + bx)^{1/2} \approx 1 + \frac{1}{2} b x \]
Thus:
\[ (1 + b x)^{1/2} - 1 \approx \frac{1}{2} b x \]
Substituting this back, we get:
\[ \lim_{x \to 0^+} f(x) = \frac{1}{b} \cdot \frac{1}{2} b x = \frac{1}{2} \cdot 0 = 0 \]

### Step 4: Set the Limits Equal for Continuity
For \( f(x) \) to be continuous at \( x = 0 \):
\[ \text{Left-hand limit} = \text{Right-hand limit} = f(0) \]
This gives us the equations:
\[ a + 2 = 0 \quad \text{(from left-hand limit)} \]
\[ c = 0 \quad \text{(from right-hand limit)} \]

### Step 5: Solve the Equations
From \( a + 2 = 0 \):
\[ a = -2 \]
From \( c = 0 \):
\[ c = 0 \]
The value of \( b \) can be any real number since it does not affect continuity.

### Final Values
Thus, the final values are:
\[ a = -2, \quad b = \text{any real number}, \quad c = 0 \]