Evaluate : `lim_(x to 0)(1-cos2x)/(3x^(2))`

Answer

To evaluate the limit \( \lim_{x \to 0} \frac{1 - \cos(2x)}{3x^2} \), we can follow these steps:

### Step 1: Identify the form of the limit
When we substitute \( x = 0 \) into the expression, we get:
\[ 1 - \cos(2 \cdot 0) = 1 - \cos(0) = 1 - 1 = 0 \]
and
\[ 3(0)^2 = 0 \]
Thus, we have the indeterminate form \( \frac{0}{0} \).

### Step 2: Apply L'Hôpital's Rule
Since we have the \( \frac{0}{0} \) form, we can apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and the derivative of the denominator separately.

#### Differentiate the numerator:
The derivative of \( 1 - \cos(2x) \) is:
\[ \frac{d}{dx}(1 - \cos(2x)) = 0 + 2\sin(2x) = 2\sin(2x) \]

#### Differentiate the denominator:
The derivative of \( 3x^2 \) is:
\[ \frac{d}{dx}(3x^2) = 6x \]

### Step 3: Rewrite the limit
Now we can rewrite the limit using the derivatives:
\[ \lim_{x \to 0} \frac{1 - \cos(2x)}{3x^2} = \lim_{x \to 0} \frac{2\sin(2x)}{6x} \]

### Step 4: Simplify the limit
We can simplify the expression:
\[ \lim_{x \to 0} \frac{2\sin(2x)}{6x} = \lim_{x \to 0} \frac{1}{3} \cdot \frac{\sin(2x)}{x} \]

### Step 5: Use the limit property
We know that:
\[ \lim_{x \to 0} \frac{\sin(kx)}{x} = k \quad \text{for any constant } k \]
In this case, \( k = 2 \). Therefore:
\[ \lim_{x \to 0} \frac{\sin(2x)}{x} = 2 \]

### Step 6: Substitute back into the limit
Now substituting this back into our limit:
\[ \lim_{x \to 0} \frac{1}{3} \cdot \frac{\sin(2x)}{x} = \frac{1}{3} \cdot 2 = \frac{2}{3} \]

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

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