Key Concepts and Formulas

• Continuity of a Function: A function $f(x)$ is continuous at a point $x=c$ if and only if $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$.
• Standard Limit: $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$: This fundamental limit is crucial for evaluating limits involving trigonometric functions.
• Limit Evaluation Techniques: For rational functions or functions involving radicals, techniques like multiplying by the conjugate or algebraic simplification are often used.

Step-by-Step Solution

Step 1: Understand the Condition for Continuity The problem states that the function $f(x)$ is continuous at $x=0$. This means that the limit of the function as $x$ approaches 0 from the left must be equal to the limit as $x$ approaches 0 from the right, and both must be equal to the value of the function at $x=0$. Mathematically, this is expressed as: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$

Step 2: Determine the Value of $f(0)$ From the definition of the function, $f(x) = b$ when $x = 0$. Therefore, $f(0) = b$.

Step 3: Evaluate the Left-Hand Limit ($\lim_{x \to 0^-} f(x)$) For $x < 0$, the function is defined as $f(x) = \frac{\sin (a + 1)x + \sin 2x}{2x}$. We need to find the limit as $x$ approaches 0 from the negative side: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{\sin (a + 1)x + \sin 2x}{2x}$ We can split this into two separate limits: $\lim_{x \to 0^-} \left( \frac{\sin (a + 1)x}{2x} + \frac{\sin 2x}{2x} \right)$ To use the standard limit $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$, we adjust the terms: $\lim_{x \to 0^-} \left( \frac{\sin (a + 1)x}{(a + 1)x} \cdot \frac{a + 1}{2} + \frac{\sin 2x}{2x} \right)$ As $x \to 0^-$, $(a+1)x \to 0$ and $2x \to 0$. Therefore, we can apply the standard limit: $\left( 1 \cdot \frac{a + 1}{2} + 1 \right) = \frac{a + 1}{2} + 1$ So, the left-hand limit is $\frac{a+1}{2} + 1$.

Step 4: Evaluate the Right-Hand Limit ($\lim_{x \to 0^+} f(x)$) For $x > 0$, the function is defined as $f(x) = \frac{\sqrt{x+b x^3}-\sqrt{x}}{b x^{5 / 2}}$. We need to find the limit as $x$ approaches 0 from the positive side: $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{\sqrt{x+b x^3}-\sqrt{x}}{b x^{5 / 2}}$ To simplify the expression, we multiply the numerator and denominator by the conjugate of the numerator, which is $\sqrt{x+b x^3}+\sqrt{x}$: $\lim_{x \to 0^+} \frac{(\sqrt{x+b x^3}-\sqrt{x})(\sqrt{x+b x^3}+\sqrt{x})}{b x^{5 / 2}(\sqrt{x+b x^3}+\sqrt{x})}$ The numerator simplifies using the difference of squares formula $(p-q)(p+q) = p^2 - q^2$: $\lim_{x \to 0^+} \frac{(x+b x^3) - x}{b x^{5 / 2}(\sqrt{x+b x^3}+\sqrt{x})}$ $\lim_{x \to 0^+} \frac{b x^3}{b x^{5 / 2}(\sqrt{x+b x^3}+\sqrt{x})}$ We can cancel out $b$ from the numerator and denominator: $\lim_{x \to 0^+} \frac{x^3}{x^{5 / 2}(\sqrt{x+b x^3}+\sqrt{x})}$ Simplify the powers of $x$: $x^3 / x^{5/2} = x^{3 - 5/2} = x^{1/2} = \sqrt{x}$. $\lim_{x \to 0^+} \frac{\sqrt{x}}{\sqrt{x+b x^3}+\sqrt{x}}$ We can factor out $\sqrt{x}$ from the denominator: $\lim_{x \to 0^+} \frac{\sqrt{x}}{\sqrt{x}(\sqrt{1+b x^2}+1)}$ Now, cancel out $\sqrt{x}$ from the numerator and denominator: $\lim_{x \to 0^+} \frac{1}{\sqrt{1+b x^2}+1}$ As $x \to 0^+$, $bx^2 \to 0$. So, the limit becomes: $\frac{1}{\sqrt{1+0}+1} = \frac{1}{1+1} = \frac{1}{2}$ So, the right-hand limit is $\frac{1}{2}$.

Step 5: Equate the Limits and $f(0)$ to Find $a$ and $b$ Since $f(x)$ is continuous at $x=0$, we have: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$ From our calculations: $\frac{a + 1}{2} + 1 = \frac{1}{2} = b$ From the equality $\frac{1}{2} = b$, we get $b = \frac{1}{2}$. Now, substitute the value of $b$ into the first equality: $\frac{a + 1}{2} + 1 = \frac{1}{2}$ Subtract 1 from both sides: $\frac{a + 1}{2} = \frac{1}{2} - 1$ $\frac{a + 1}{2} = -\frac{1}{2}$ Multiply both sides by 2: $a + 1 = -1$ Subtract 1 from both sides: $a = -2$

Step 6: Calculate $a+b$ We have found that $a = -2$ and $b = \frac{1}{2}$. The problem asks for the value of $a+b$. $a + b = -2 + \frac{1}{2}$ To add these, find a common denominator: $a + b = -\frac{4}{2} + \frac{1}{2}$ $a + b = \frac{-4 + 1}{2}$ $a + b = -\frac{3}{2}$

Common Mistakes & Tips

• Algebraic Errors: Be meticulous with algebraic manipulations, especially when dealing with fractions, exponents, and square roots. A small error can lead to an incorrect final answer.
• Misapplication of Standard Limits: Ensure that the argument of the sine function in the numerator matches the denominator precisely when applying $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$. If there's a constant multiplier, adjust accordingly.
• Conjugate Multiplication: When rationalizing a term with two square roots, always multiply both the numerator and the denominator by the conjugate to maintain the value of the expression.

Summary

The problem requires us to use the definition of continuity at a point, which states that the left-hand limit, the right-hand limit, and the function's value at that point must all be equal. We evaluated the left-hand limit using the standard sine limit formula, the right-hand limit by rationalizing the numerator, and identified $f(0)$ directly from the function definition. By equating these values, we derived the values of $a$ and $b$, and subsequently computed their sum.

The final answer is $\boxed{- \frac{3}{2}}$.