`lim_(xrarr0) (sin4x-sin2x+x)/(x)=?`

Options

A. 3
B. 4
C. 6
D. 7

Answer

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

### Step 1: Identify the Form
First, we substitute \( x = 0 \) into the expression:
\[ \sin(4 \cdot 0) - \sin(2 \cdot 0) + 0 = 0 - 0 + 0 = 0 \]
Thus, we have a \( \frac{0}{0} \) indeterminate form.

### Step 2: Apply L'Hôpital's Rule
Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that if \( \lim_{x \to c} \frac{f(x)}{g(x)} = \frac{0}{0} \) or \( \frac{\infty}{\infty} \), then:
\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \]
Here, \( f(x) = \sin(4x) - \sin(2x) + x \) and \( g(x) = x \).

### Step 3: Differentiate the Numerator and Denominator
Now, we differentiate the numerator and the denominator:
- The derivative of the numerator \( f(x) \):
\[ f'(x) = 4\cos(4x) - 2\cos(2x) + 1 \]
- The derivative of the denominator \( g(x) \):
\[ g'(x) = 1 \]

### Step 4: Evaluate the New Limit
Now we evaluate the limit:
\[ \lim_{x \to 0} \frac{f'(x)}{g'(x)} = \lim_{x \to 0} (4\cos(4x) - 2\cos(2x) + 1) \]
Substituting \( x = 0 \):
\[ = 4\cos(0) - 2\cos(0) + 1 = 4 \cdot 1 - 2 \cdot 1 + 1 = 4 - 2 + 1 = 3 \]

### Final Answer
Thus, the limit is:
\[ \lim_{x \to 0} \frac{\sin 4x - \sin 2x + x}{x} = 3 \]