If `lim_(xrarr0)(2a sin x-sin 2x)/(tan^(3)x)` exists and is equal to 1, then the value of a is

Options

A. 2
B. 1
C. 0
D. `-1`

Answer

To solve the limit problem given, we need to evaluate the limit:

\[ \lim_{x \to 0} \frac{2a \sin x - \sin 2x}{\tan^3 x} \]

and find the value of \( a \) such that this limit equals 1.

### Step 1: Rewrite the limit
We start with the limit expression:

\[ L = \lim_{x \to 0} \frac{2a \sin x - \sin 2x}{\tan^3 x} \]

### Step 2: Use the identity for \(\sin 2x\)
Recall that \(\sin 2x = 2 \sin x \cos x\). Thus, we can rewrite the limit as:

\[ L = \lim_{x \to 0} \frac{2a \sin x - 2 \sin x \cos x}{\tan^3 x} \]

### Step 3: Factor out \(2 \sin x\)
Factoring out \(2 \sin x\) from the numerator gives:

\[ L = \lim_{x \to 0} \frac{2 \sin x (a - \cos x)}{\tan^3 x} \]

### Step 4: Rewrite \(\tan x\)
We know that \(\tan x = \frac{\sin x}{\cos x}\), therefore \(\tan^3 x = \frac{\sin^3 x}{\cos^3 x}\). Substituting this in gives:

\[ L = \lim_{x \to 0} \frac{2 \sin x (a - \cos x) \cos^3 x}{\sin^3 x} \]

### Step 5: Simplify the expression
This simplifies to:

\[ L = \lim_{x \to 0} 2(a - \cos x) \frac{\cos^3 x}{\sin^2 x} \]

### Step 6: Evaluate the limit
As \(x\) approaches 0, \(\cos x\) approaches 1, and we can use the limit:

\[ \lim_{x \to 0} \frac{a - \cos x}{x^2} = \frac{1}{2} \]

This is a standard limit, which states that:

\[ \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2} \]

### Step 7: Set up the equation
Thus, we have:

\[ L = 2 \cdot \lim_{x \to 0} \frac{a - \cos x}{\sin^2 x} \cdot \cos^3(0) \]

Since \(\cos(0) = 1\):

\[ L = 2 \cdot \lim_{x \to 0} \frac{a - \cos x}{x^2} \]

### Step 8: Set the limit equal to 1
We know that:

\[ \lim_{x \to 0} \frac{a - \cos x}{x^2} = \frac{1}{2} \]

Thus:

\[ 2 \cdot \frac{1}{2} = 1 \]

### Step 9: Solve for \(a\)
From the limit, we have:

\[ a - \cos x \approx \frac{1}{2} x^2 \]

At \(x = 0\), \(\cos 0 = 1\):

\[ a - 1 = 0 \implies a = 1 \]

### Conclusion
Thus, the value of \(a\) is:

\[ \boxed{1} \]