Key Concepts and Formulas

• Continuity at a point: A function $f(x)$ is continuous at a point $x=c$ if and only if the following three conditions are met:

  1. $f(c)$ is defined.
  2. $\lim_{x \to c} f(x)$ exists.
  3. $\lim_{x \to c} f(x) = f(c)$. For a piecewise function, this means the left-hand limit, the right-hand limit, and the function value at the point must all be equal. $\lim_{x \to c^{-}} f(x) = \lim_{x \to c^{+}} f(x) = f(c)$.

• Standard Limits:

  • $\lim_{x \to 0} \frac{\sin x}{x} = 1$
  • $\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$
  • $\lim_{x \to 0} \frac{1 - \cos(kx)}{x^2} = \frac{k^2}{2}$

• Trigonometric Identity: $1 - \cos(2\theta) = 2\sin^2(\theta)$.

Step-by-Step Solution

Step 1: Understand the condition for continuity. The problem states that the function $f(x)$ is continuous at $x=0$. For a function to be continuous at a point $x=c$, the left-hand limit, the right-hand limit, and the function value at that point must be equal. Therefore, we must have: $\lim_{x \to 0^{-}} f(x) = \lim_{x \to 0^{+}} f(x) = f(0)$

Step 2: Determine the value of $f(0)$. From the definition of the function, $f(x) = \alpha$ when $x=0$. So, $f(0) = \alpha$.

Step 3: Calculate the left-hand limit, $\lim_{x \to 0^{-}} f(x)$. For $x < 0$, the function is defined as $f(x) = \frac{1 - \cos 2x}{x^2}$. We need to find the limit of this expression as $x$ approaches 0 from the left. $\lim_{x \to 0^{-}} f(x) = \lim_{x \to 0^{-}} \frac{1 - \cos 2x}{x^2}$ We can use the standard limit $\lim_{x \to 0} \frac{1 - \cos(kx)}{x^2} = \frac{k^2}{2}$. Here, $k=2$. $\lim_{x \to 0^{-}} \frac{1 - \cos 2x}{x^2} = \frac{2^2}{2} = \frac{4}{2} = 2$ Alternatively, using the identity $1 - \cos(2x) = 2\sin^2(x)$: $\lim_{x \to 0^{-}} \frac{2\sin^2 x}{x^2} = 2 \lim_{x \to 0^{-}} \left(\frac{\sin x}{x}\right)^2$ Since $\lim_{x \to 0} \frac{\sin x}{x} = 1$, we have: $2 \times (1)^2 = 2$ So, $\lim_{x \to 0^{-}} f(x) = 2$.

Step 4: Equate the left-hand limit and $f(0)$ to find $\alpha$. From Step 1, $\lim_{x \to 0^{-}} f(x) = f(0)$. Therefore, $2 = \alpha$.

Step 5: Calculate the right-hand limit, $\lim_{x \to 0^{+}} f(x)$. For $x > 0$, the function is defined as $f(x) = \frac{\beta \sqrt{1 - \cos x}}{x}$. We need to find the limit of this expression as $x$ approaches 0 from the right. $\lim_{x \to 0^{+}} f(x) = \lim_{x \to 0^{+}} \frac{\beta \sqrt{1 - \cos x}}{x}$ We can use the identity $1 - \cos x = 2\sin^2\left(\frac{x}{2}\right)$. $\lim_{x \to 0^{+}} \frac{\beta \sqrt{2\sin^2\left(\frac{x}{2}\right)}}{x}$ Since $x > 0$, $\frac{x}{2} > 0$, and for small positive values, $\sin\left(\frac{x}{2}\right) > 0$. Thus, $\sqrt{\sin^2\left(\frac{x}{2}\right)} = \sin\left(\frac{x}{2}\right)$. $\lim_{x \to 0^{+}} \frac{\beta \sqrt{2} \sin\left(\frac{x}{2}\right)}{x}$ To use the standard limit $\lim_{y \to 0} \frac{\sin y}{y} = 1$, we can rewrite the expression: $\lim_{x \to 0^{+}} \frac{\beta \sqrt{2}}{2} \cdot \frac{\sin\left(\frac{x}{2}\right)}{\frac{x}{2}}$ Let $y = \frac{x}{2}$. As $x \to 0^{+}$, $y \to 0^{+}$. So, $\lim_{x \to 0^{+}} \frac{\sin\left(\frac{x}{2}\right)}{\frac{x}{2}} = \lim_{y \to 0^{+}} \frac{\sin y}{y} = 1$. $\frac{\beta \sqrt{2}}{2} \times 1 = \frac{\beta}{\sqrt{2}}$ So, $\lim_{x \to 0^{+}} f(x) = \frac{\beta}{\sqrt{2}}$.

Step 6: Equate the right-hand limit and $f(0)$ to find $\beta$. From Step 1, $\lim_{x \to 0^{+}} f(x) = f(0)$. We found $f(0) = \alpha = 2$ from Step 4, and $\lim_{x \to 0^{+}} f(x) = \frac{\beta}{\sqrt{2}}$ from Step 5. Therefore, $2 = \frac{\beta}{\sqrt{2}}$. Solving for $\beta$: $\beta = 2\sqrt{2}$

Step 7: Calculate $\alpha^2 + \beta^2$. We have found $\alpha = 2$ and $\beta = 2\sqrt{2}$. Now, we compute $\alpha^2 + \beta^2$: $\alpha^2 = (2)^2 = 4$ $\beta^2 = (2\sqrt{2})^2 = 4 \times 2 = 8$ $\alpha^2 + \beta^2 = 4 + 8 = 12$

Common Mistakes & Tips

• Sign errors with square roots: When dealing with $\sqrt{x^2}$, remember that $\sqrt{x^2} = |x|$. In the case of $\sqrt{\sin^2(x/2)}$, since we are considering $x \to 0^{+}$, $x/2$ is positive, so $\sin(x/2)$ is positive, and $\sqrt{\sin^2(x/2)} = \sin(x/2)$.
• Incorrect application of standard limits: Ensure that the argument of the trigonometric function matches the denominator. For example, in $\lim_{x \to 0} \frac{\sin(x/2)}{x}$, you need to manipulate it to $\lim_{x \to 0} \frac{\sin(x/2)}{x/2} \cdot \frac{1}{2}$.
• Forgetting to equate all three parts: For continuity, $f(0)$, the left-hand limit, and the right-hand limit must all be equal. Do not just equate the limits without considering $f(0)$.

Summary

To ensure continuity of the function $f(x)$ at $x=0$, we must equate the function's value at $x=0$ with both the left-hand limit and the right-hand limit as $x$ approaches 0. By evaluating the left-hand limit $\lim_{x \to 0^{-}} f(x) = \frac{1-\cos 2x}{x^2}$, we found it to be 2. This implies $\alpha = 2$. Next, by evaluating the right-hand limit $\lim_{x \to 0^{+}} f(x) = \frac{\beta \sqrt{1-\cos x}}{x}$, we found it to be $\frac{\beta}{\sqrt{2}}$. Equating this to $\alpha=2$, we get $\frac{\beta}{\sqrt{2}} = 2$, which yields $\beta = 2\sqrt{2}$. Finally, we calculated $\alpha^2 + \beta^2 = 2^2 + (2\sqrt{2})^2 = 4 + 8 = 12$.

The final answer is $\boxed{12}$.