`lim_(x to 0) (1- cos 2x)/x`

Options

A. 0
B. 1
C. 2
D. 4

Answer

To solve the limit \( \lim_{x \to 0} \frac{1 - \cos(2x)}{x} \), we can follow these steps:

### Step 1: Substitute \( x = 0 \)
First, we substitute \( x = 0 \) directly into the limit:
\[ \frac{1 - \cos(2 \cdot 0)}{0} = \frac{1 - \cos(0)}{0} = \frac{1 - 1}{0} = \frac{0}{0} \]
This gives us an indeterminate form \( \frac{0}{0} \).

**Hint:** If you encounter \( \frac{0}{0} \) when evaluating a limit, consider using L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule
Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that:
\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \]
if the limit on the right side exists.

Here, let \( f(x) = 1 - \cos(2x) \) and \( g(x) = x \). We need to find the derivatives of the numerator and the denominator.

**Hint:** Differentiate the numerator and denominator separately.

### Step 3: Differentiate the Numerator and Denominator
1. **Numerator:**
\[ f'(x) = \frac{d}{dx}(1 - \cos(2x)) = 0 + 2\sin(2x) = 2\sin(2x) \]
2. **Denominator:**
\[ g'(x) = \frac{d}{dx}(x) = 1 \]

Now we can rewrite the limit using these derivatives:
\[ \lim_{x \to 0} \frac{f'(x)}{g'(x)} = \lim_{x \to 0} \frac{2\sin(2x)}{1} \]

**Hint:** After applying L'Hôpital's Rule, substitute \( x = 0 \) again.

### Step 4: Evaluate the New Limit
Now we evaluate the new limit:
\[ \lim_{x \to 0} 2\sin(2x) = 2\sin(0) = 2 \cdot 0 = 0 \]

Thus, the final result is:
\[ \lim_{x \to 0} \frac{1 - \cos(2x)}{x} = 0 \]

### Final Answer
The limit is \( 0 \).

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