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: Rewrite the limit using the double angle formula
We start by using the double angle formula for sine:
\[ \sin 2x = 2 \sin x \cos x \]
Thus, we can rewrite the limit as:
\[ \lim_{x \to 0} \frac{2 \sin x \cos x - a \sin x}{x^3} \]

### Step 2: Factor out \( \sin x \)
Next, we can factor out \( \sin x \) from the numerator:
\[ \lim_{x \to 0} \frac{\sin x (2 \cos x - a)}{x^3} \]

### Step 3: Rewrite \( x^3 \) in terms of \( \sin x \)
We can express \( x^3 \) as \( x^2 \cdot x \) and use the fact that \( \frac{\sin x}{x} \to 1 \) as \( x \to 0 \):
\[ \lim_{x \to 0} \frac{\sin x}{x} \cdot \frac{2 \cos x - a}{x^2} \]

### Step 4: Evaluate the limit
Now, we know that:
\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \]
So we need to focus on the limit:
\[ \lim_{x \to 0} \frac{2 \cos x - a}{x^2} \]

### Step 5: Determine the condition for the limit to exist
For the limit to exist finitely, the numerator \( 2 \cos x - a \) must approach 0 as \( x \to 0 \). Evaluating at \( x = 0 \):
\[ 2 \cos(0) - a = 2 - a \]
Setting this equal to 0 gives:
\[ 2 - a = 0 \implies a = 2 \]

### Step 6: Verify the limit with \( a = 2 \)
Substituting \( a = 2 \) back into the limit:
\[ \lim_{x \to 0} \frac{2 \cos x - 2}{x^2} = \lim_{x \to 0} \frac{2(\cos x - 1)}{x^2} \]
Using the Taylor series expansion for \( \cos x \) around \( x = 0 \):
\[ \cos x \approx 1 - \frac{x^2}{2} + O(x^4) \]
Thus:
\[ \cos x - 1 \approx -\frac{x^2}{2} \]
Substituting this into our limit:
\[ \lim_{x \to 0} \frac{2 \left(-\frac{x^2}{2}\right)}{x^2} = \lim_{x \to 0} -1 = -1 \]
This shows that the limit exists and is finite when \( a = 2 \).

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