Question:

Step 1: Establish the condition for continuity For $f(x)$ to be continuous at $x=0$, the limit of $f(x)$ as $x \to 0$ must exist and be equal to $f(0)$. $$ \lim_{x \to 0} f(x) = f(0) $$ Step 2: Expand the numerator using Taylor series The Taylor series expansions for $\sin x$ and $\sin 2x$ around $x=0$ are: $$ \sin x = x - \frac{x^3}{3!} + O(x^5) = x - \frac{x^3}{6} + O(x^5) $$ $$ \sin 2x = (2x) - \frac{(2x)^3}{3!} + O(x^5) = 2x - \frac{8x^3}{6} + O(x^5) = 2x - \frac{4x^3}{3} + O(x^5) $$ Substitute these into the numerator of $f(x)$: $$ A \sin x + \sin 2x = A\left(x - \frac{x^3}{6} + O(x^5)\right) + \left(2x - \frac{4x^3}{3} + O(x^5)\right) $$ $$ A \sin x + \sin 2x = (A+2)x + \left(-\frac{A}{6} - \frac{4}{3}\right)x^3 + O(x^5) $$ Step 3: Determine the value of A for continuity The function is given by: $$ f(x) = \frac{(A+2)x + \left(-\frac{A}{6} - \frac{4}{3}\right)x^3 + O(x^5)}{x^3} $$ For $\lim_{x \to 0} f(x)$ to exist and be finite, the term containing $x$ in the numerator must vanish. Otherwise, the expression would contain a term proportional to $\frac{x}{x^3} = \frac{1}{x^2}$, which diverges as $x \to 0$. Therefore, the coefficient of $x$ in the numerator must be zero: $$ A+2 = 0 $$ $$ A = -2 $$ Step 4: Calculate f(0) Substitute $A = -2$ into the numerator: $$ A \sin x + \sin 2x = (-2+2)x + \left(-\frac{-2}{6} - \frac{4}{3}\right)x^3 + O(x^5) $$ $$ A \sin x + \sin 2x = 0 \cdot x + \left(\frac{1}{3} - \frac{4}{3}\right)x^3 + O(x^5) $$ $$ A \sin x + \sin 2x = -x^3 + O(x^5) $$ Now, substitute this back into $f(x)$: $$ f(x) = \frac{-x^3 + O(x^5)}{x^3} = -1 + O(x^2) $$ Taking the limit as $x \to 0$: $$ \lim_{x \to 0} f(x) = \lim_{x \to 0} (-1 + O(x^2)) = -1 $$ For continuity, $f(0)$ must be equal to this limit. $$ f(0) = -1 $$