`lim_(xrarr0) (3 tan3x-4 tan2x-tanx)/(4x^(2)tanx)`

Options

A. 0
B. 1
C. 3
D. 4

Explanation

To solve the limit \( \lim_{x \to 0} \frac{3 \tan(3x) - 4 \tan(2x) - \tan(x)}{4x^2 \tan(x)} \), we will follow these steps:

### Step 1: Rewrite the limit
We start with the limit expression:
\[ \lim_{x \to 0} \frac{3 \tan(3x) - 4 \tan(2x) - \tan(x)}{4x^2 \tan(x)} \]

### Step 2: Use Taylor series expansion for \(\tan(kx)\)
Recall the Taylor series expansion for \(\tan(x)\) around \(x = 0\):
\[ \tan(kx) \approx kx + \frac{(kx)^3}{3} + O(x^5) \]
Using this, we can expand \(\tan(3x)\), \(\tan(2x)\), and \(\tan(x)\):
- \(\tan(3x) \approx 3x + \frac{(3x)^3}{3} = 3x + 9x^3\)
- \(\tan(2x) \approx 2x + \frac{(2x)^3}{3} = 2x + \frac{8x^3}{3}\)
- \(\tan(x) \approx x + \frac{x^3}{3}\)

### Step 3: Substitute the expansions into the limit
Substituting these expansions into our limit gives:
\[ 3(3x + 9x^3) - 4(2x + \frac{8x^3}{3}) - (x + \frac{x^3}{3}) \]
Calculating this:
\[ = 9x + 27x^3 - 8x - \frac{32x^3}{3} - x - \frac{x^3}{3} \]
Combining like terms:
\[ = (9x - 8x - x) + \left(27x^3 - \frac{32x^3}{3} - \frac{x^3}{3}\right) \]
\[ = 0 + \left(27x^3 - \frac{33x^3}{3}\right) = 0 + \left(27x^3 - 11x^3\right) = 16x^3 \]

### Step 4: Substitute back into the limit
Now we substitute back into the limit:
\[ \lim_{x \to 0} \frac{16x^3}{4x^2 \tan(x)} \]
Using the expansion \(\tan(x) \approx x\):
\[ = \lim_{x \to 0} \frac{16x^3}{4x^2 \cdot x} = \lim_{x \to 0} \frac{16x^3}{4x^3} = \lim_{x \to 0} \frac{16}{4} = 4 \]

### Final Answer
Thus, the limit evaluates to:
\[ \boxed{4} \]