Evaluate the following Limits
`lim_(xto 0)(1-cos2x)/(x*tan3x)`

Answer

To evaluate the limit

\[ \lim_{x \to 0} \frac{1 - \cos(2x)}{x \tan(3x)}, \]

we will follow these steps:

### Step 1: Substitute \( x = 0 \)

First, we substitute \( x = 0 \) directly into the limit:

\[ \frac{1 - \cos(2 \cdot 0)}{0 \cdot \tan(3 \cdot 0)} = \frac{1 - \cos(0)}{0 \cdot \tan(0)} = \frac{1 - 1}{0} = \frac{0}{0}. \]

This is an indeterminate form, so we need to simplify the expression.

**Hint:** When you encounter a \( \frac{0}{0} \) form, consider using trigonometric identities or L'Hôpital's Rule.

### Step 2: Use the trigonometric identity

We use the identity \( 1 - \cos(2x) = 2 \sin^2(x) \):

\[ \lim_{x \to 0} \frac{1 - \cos(2x)}{x \tan(3x)} = \lim_{x \to 0} \frac{2 \sin^2(x)}{x \tan(3x)}. \]

**Hint:** Remember that \( \tan(3x) = \frac{\sin(3x)}{\cos(3x)} \).

### Step 3: Rewrite the limit

Now we can rewrite the limit using the definition of tangent:

\[ \lim_{x \to 0} \frac{2 \sin^2(x)}{x \cdot \frac{\sin(3x)}{\cos(3x)}} = \lim_{x \to 0} \frac{2 \sin^2(x) \cos(3x)}{x \sin(3x)}. \]

**Hint:** It may help to separate the limit into parts that can be evaluated individually.

### Step 4: Split the limit

We can separate the limit into two parts:

\[ \lim_{x \to 0} \frac{2 \sin^2(x)}{x} \cdot \lim_{x \to 0} \frac{\cos(3x)}{\sin(3x)}. \]

**Hint:** Use the known limits \( \lim_{x \to 0} \frac{\sin(x)}{x} = 1 \) and \( \lim_{x \to 0} \cos(x) = 1 \).

### Step 5: Evaluate the limits

1. For the first limit, we can rewrite \( \sin^2(x) \) as \( \left(\frac{\sin(x)}{x}\right)^2 \cdot x^2 \):

\[ \lim_{x \to 0} \frac{2 \sin^2(x)}{x} = 2 \cdot \lim_{x \to 0} \frac{\sin^2(x)}{x^2} \cdot x = 2 \cdot 1 \cdot 0 = 0. \]

2. For the second limit:

\[ \lim_{x \to 0} \frac{\cos(3x)}{\sin(3x)} = \lim_{x \to 0} \frac{1}{3} \cdot \frac{\cos(3x)}{x} = \frac{1}{3} \cdot 1 = \frac{1}{3}. \]

**Hint:** Remember that \( \sin(3x) \) can be approximated as \( 3x \) when \( x \) is near 0.

### Step 6: Combine the results

Now we combine the results:

\[ \lim_{x \to 0} \frac{2 \sin^2(x)}{x} \cdot \lim_{x \to 0} \frac{\cos(3x)}{\sin(3x)} = 0 \cdot \frac{1}{3} = 0. \]

Thus, the final result is:

\[ \boxed{0}. \]