If `lim_(xrarr0)(sin2x-asinx)/(x^(3))` exists finitely, then the value of a is

Options

A. 0
B. 2
C. 1
D. 4

Answer

To solve the limit problem \( \lim_{x \to 0} \frac{\sin 2x - a \sin x}{x^3} \) and find the value of \( a \) for which this limit exists finitely, we can follow these steps:

### Step 1: Expand \( \sin 2x \) and \( \sin x \) using Taylor series
The Taylor series expansion for \( \sin x \) around \( x = 0 \) is:
\[ \sin x = x - \frac{x^3}{6} + O(x^5) \]
For \( \sin 2x \), we can substitute \( 2x \) into the series:
\[ \sin 2x = 2x - \frac{(2x)^3}{6} + O(x^5) = 2x - \frac{8x^3}{6} + O(x^5) = 2x - \frac{4x^3}{3} + O(x^5) \]

### Step 2: Substitute the expansions into the limit
Now substituting the expansions into the limit expression:
\[ \sin 2x - a \sin x = \left(2x - \frac{4x^3}{3} + O(x^5)\right) - a\left(x - \frac{x^3}{6} + O(x^5)\right) \]
This simplifies to:
\[ = 2x - \frac{4x^3}{3} - ax + \frac{ax^3}{6} + O(x^5) \]
Combining like terms:
\[ = (2 - a)x + \left(-\frac{4}{3} + \frac{a}{6}\right)x^3 + O(x^5) \]

### Step 3: Formulate the limit expression
Now, we can write the limit:
\[ \lim_{x \to 0} \frac{(2 - a)x + \left(-\frac{4}{3} + \frac{a}{6}\right)x^3 + O(x^5)}{x^3} \]
This can be separated into two parts:
\[ = \lim_{x \to 0} \left( \frac{(2 - a)x}{x^3} + \frac{-\frac{4}{3} + \frac{a}{6}}{1} + O(x^2) \right) \]
The first term \(\frac{(2 - a)x}{x^3}\) simplifies to \(\frac{2 - a}{x^2}\), which goes to infinity unless \(2 - a = 0\).

### Step 4: Set the condition for the limit to exist
For the limit to exist finitely, we need:
\[ 2 - a = 0 \implies a = 2 \]

### Final Answer
Thus, the value of \( a \) for which the limit exists finitely is:
\[ \boxed{2} \]