Key Concepts and Formulas

• Continuity of a Function: A function $f(x)$ is continuous at a point $x=c$ if $\lim_{x \to c} f(x) = f(c)$.
• Taylor Series Expansion: For small values of $x$, the Taylor series expansions of trigonometric functions are:
  • $\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots$
  • $\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots$
• Limit Properties: The limit of a sum/difference is the sum/difference of the limits, and the limit of a product is the product of the limits, provided these limits exist. Also, $\lim_{x \to 0} \frac{\sin x}{x} = 1$.

Step-by-Step Solution

Step 1: Apply the definition of continuity. The problem states that the function $f(x)$ is continuous at $x=0$. By the definition of continuity, this means that the limit of the function as $x$ approaches 0 must be equal to the value of the function at $x=0$. $\lim_{x \to 0} f(x) = f(0)$ Substituting the given function, we have: $\lim_{x \to 0} \frac{\sin 3x + \alpha \sin x - \beta \cos 3x}{x^3} = f(0)$ For the limit on the left-hand side to exist, the numerator must approach zero as $x \to 0$ if the denominator approaches zero.

Step 2: Analyze the numerator as $x \to 0$. Let's evaluate the terms in the numerator as $x \to 0$:

• $\sin 3x \to \sin(0) = 0$
• $\alpha \sin x \to \alpha \sin(0) = 0$
• $\beta \cos 3x \to \beta \cos(0) = \beta \times 1 = \beta$

For the limit $\lim_{x \to 0} \frac{\sin 3x + \alpha \sin x - \beta \cos 3x}{x^3}$ to exist, the numerator must approach 0 as $x \to 0$ because the denominator $x^3$ approaches 0. Therefore, the constant term in the numerator must be zero. $0 + 0 - \beta = 0 \implies \beta = 0$

Step 3: Simplify the expression with $\beta = 0$. Now that we know $\beta = 0$, the expression for the limit becomes: $\lim_{x \to 0} \frac{\sin 3x + \alpha \sin x}{x^3}$ For this limit to exist, the numerator must still approach zero as $x \to 0$. We already know $\sin 3x \to 0$ and $\alpha \sin x \to 0$. This condition is satisfied. However, to evaluate the limit, we need to consider the behavior of the numerator up to the $x^3$ term.

Step 4: Use Taylor series expansions for the numerator. We use the Taylor series expansion of $\sin x$ around $x=0$: $\sin x = x - \frac{x^3}{6} + O(x^5)$. Applying this to $\sin 3x$ and $\sin x$:

• $\sin 3x = (3x) - \frac{(3x)^3}{6} + O(x^5) = 3x - \frac{27x^3}{6} + O(x^5)$
• $\sin x = x - \frac{x^3}{6} + O(x^5)$

Substitute these into the numerator: $\sin 3x + \alpha \sin x = \left(3x - \frac{27x^3}{6} + O(x^5)\right) + \alpha \left(x - \frac{x^3}{6} + O(x^5)\right)$ $= (3x + \alpha x) - \left(\frac{27}{6} + \frac{\alpha}{6}\right)x^3 + O(x^5)$ $= (3+\alpha)x - \frac{27+\alpha}{6}x^3 + O(x^5)$

Now, the limit becomes: $\lim_{x \to 0} \frac{(3+\alpha)x - \frac{27+\alpha}{6}x^3 + O(x^5)}{x^3}$

Step 5: Analyze the limit for existence. For the limit to exist as $x \to 0$, the terms with powers of $x$ less than 3 in the numerator must cancel out. The term $(3+\alpha)x$ is a first-order term in $x$. If $3+\alpha \neq 0$, then as $x \to 0$, this term will dominate and the limit will be $\lim_{x \to 0} \frac{(3+\alpha)x}{x^3} = \lim_{x \to 0} \frac{3+\alpha}{x^2}$, which tends to infinity. Therefore, for the limit to be finite, the coefficient of $x$ must be zero. $3 + \alpha = 0 \implies \alpha = -3$

Step 6: Evaluate the limit with $\alpha = -3$. Substitute $\alpha = -3$ back into the expression for the numerator: The term with $x$ becomes $(3+(-3))x = 0x = 0$. The term with $x^3$ becomes $-\frac{27+(-3)}{6}x^3 = -\frac{24}{6}x^3 = -4x^3$.

So, the numerator is approximately $-4x^3$ for small $x$. The limit expression now is: $\lim_{x \to 0} \frac{-4x^3 + O(x^5)}{x^3}$ We can split this limit: $\lim_{x \to 0} \left(\frac{-4x^3}{x^3} + \frac{O(x^5)}{x^3}\right) = \lim_{x \to 0} (-4 + O(x^2))$ As $x \to 0$, $O(x^2) \to 0$. $\lim_{x \to 0} (-4 + O(x^2)) = -4$

Step 7: Determine $f(0)$. Since $f(x)$ is continuous at $x=0$, we have $f(0) = \lim_{x \to 0} f(x)$. From Step 6, we found that $\lim_{x \to 0} f(x) = -4$. Therefore, $f(0) = -4$.

Alternative Approach using L'Hôpital's Rule:

If we assume the limit exists, we can use L'Hôpital's Rule. $\lim_{x \to 0} \frac{\sin 3x + \alpha \sin x - \beta \cos 3x}{x^3}$ As shown in Step 2, for the limit to exist, $\beta=0$. $\lim_{x \to 0} \frac{\sin 3x + \alpha \sin x}{x^3}$ This is of the form $\frac{0}{0}$. Applying L'Hôpital's Rule once: $\lim_{x \to 0} \frac{3 \cos 3x + \alpha \cos x}{3x^2}$ For this limit to exist, the numerator must approach 0 as $x \to 0$. $3 \cos 0 + \alpha \cos 0 = 3(1) + \alpha(1) = 3 + \alpha$ So, $3 + \alpha = 0 \implies \alpha = -3$.

Now, substitute $\alpha = -3$ and apply L'Hôpital's Rule again: $\lim_{x \to 0} \frac{3 \cos 3x - 3 \cos x}{3x^2}$ $= \lim_{x \to 0} \frac{\cos 3x - \cos x}{x^2}$ This is still of the form $\frac{0}{0}$. Applying L'Hôpital's Rule a third time: $\lim_{x \to 0} \frac{-3 \sin 3x + \sin x}{2x}$ This is still of the form $\frac{0}{0}$. Applying L'Hôpital's Rule a fourth time: $\lim_{x \to 0} \frac{-9 \cos 3x + \cos x}{2}$ Now, substitute $x=0$: $\frac{-9 \cos 0 + \cos 0}{2} = \frac{-9(1) + 1}{2} = \frac{-8}{2} = -4$ So, $\lim_{x \to 0} f(x) = -4$. Since $f(x)$ is continuous at $x=0$, $f(0) = \lim_{x \to 0} f(x) = -4$.

Common Mistakes & Tips

• Forgetting the condition for limit existence: When dealing with limits of the form $\frac{g(x)}{h(x)}$ where $h(x) \to 0$, it is crucial that $g(x) \to 0$ as well for the limit to exist as a finite value. This is what leads to determining $\beta=0$ and $3+\alpha=0$.
• Incorrect Taylor Series Expansion: Ensure the Taylor series expansions for $\sin x$ and $\cos x$ are recalled correctly, especially the powers of $x$ and the factorials in the denominators.
• Algebraic Errors: Be careful with algebraic manipulations, especially when combining terms and simplifying fractions.
• L'Hôpital's Rule application: While L'Hôpital's rule can be used, it requires repeated differentiation and can be prone to errors if not applied carefully. The Taylor series method is often more direct for polynomial or series expansions.

Summary

The problem requires finding $f(0)$ for a function $f(x)$ that is continuous at $x=0$. The definition of continuity states that $\lim_{x \to 0} f(x) = f(0)$. We evaluated the limit of the given function as $x \to 0$. For the limit to exist, the numerator must approach zero as $x \to 0$. This condition allowed us to determine that $\beta=0$. Then, using Taylor series expansions for $\sin 3x$ and $\sin x$, we expressed the numerator in terms of powers of $x$. For the limit to be finite, the coefficients of the lower-order terms in $x$ must be zero, which led to $\alpha=-3$. Substituting these values, the limit simplified to a constant, which is the value of $f(0)$.

The final answer is \boxed{4}.