`Lt_(x to0)(tan3x-2x)/(3x-sin^(2)x)` is equal to

Options

A. `1/3`
B. `-1/3`
C. `1/2`
D. `-1/2`

Answer

To solve the limit \( \lim_{x \to 0} \frac{\tan(3x) - 2x}{3x - \sin^2(x)} \), we will follow these steps:

### Step 1: Rewrite the limit
We start by rewriting the limit to make it easier to evaluate:
\[ \lim_{x \to 0} \frac{\tan(3x) - 2x}{3x - \sin^2(x)} = \lim_{x \to 0} \frac{\frac{\tan(3x)}{x} - 2}{3 - \frac{\sin^2(x)}{x}} \]

### Step 2: Divide by \( x \)
We divide both the numerator and the denominator by \( x \):
\[ = \lim_{x \to 0} \frac{\frac{\tan(3x)}{x} - 2}{3 - \frac{\sin^2(x)}{x}} \]

### Step 3: Evaluate the limit
Now we need to evaluate this limit. As \( x \to 0 \), we can use the Taylor series expansions or derivatives to find the limits of the functions involved.

1. The limit of \( \frac{\tan(3x)}{x} \) as \( x \to 0 \):
\[ \tan(3x) \approx 3x + \frac{(3x)^3}{3} = 3x + 9x^3 \text{ (using Taylor series)} \]
Thus,
\[ \frac{\tan(3x)}{x} \approx 3 + 9x^2 \to 3 \text{ as } x \to 0 \]

2. The limit of \( \frac{\sin^2(x)}{x} \) as \( x \to 0 \):
\[ \sin(x) \approx x - \frac{x^3}{6} \text{ (using Taylor series)} \]
Thus,
\[ \sin^2(x) \approx x^2 \text{ and } \frac{\sin^2(x)}{x} \approx x \to 0 \text{ as } x \to 0 \]

### Step 4: Substitute the limits
Substituting these limits back into our expression:
\[ = \frac{3 - 2}{3 - 0} = \frac{1}{3} \]

### Final Answer
Thus, the limit is:
\[ \lim_{x \to 0} \frac{\tan(3x) - 2x}{3x - \sin^2(x)} = \frac{1}{3} \]