Evaluate the following limits :
`Lim_(x to 0) (sin 2x (1- cos 2x))/(x^(3))`

Answer

To evaluate the limit

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

we can follow these steps:

### Step 1: Rewrite the expression

We can separate the limit into two parts:

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

### Step 2: Use the identity for \(1 - \cos(2x)\)

Recall the trigonometric identity:

\[ 1 - \cos(2x) = 2\sin^2(x). \]

Substituting this into our limit gives:

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

### Step 3: Simplify the first limit

For the first limit, we can rewrite \(\sin(2x)\) as:

\[ \sin(2x) = 2\sin(x)\cos(x). \]

Thus, we have:

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

Using the known limit \(\lim_{x \to 0} \frac{\sin(x)}{x} = 1\) and \(\lim_{x \to 0} \cos(x) = 1\), we find:

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

### Step 4: Simplify the second limit

Now, for the second limit:

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

Again using the known limit:

\[ \lim_{x \to 0} \left(\frac{\sin(x)}{x}\right) = 1, \]

we find:

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

### Step 5: Combine the results

Now we combine both limits:

\[ \lim_{x \to 0} \frac{\sin(2x)}{x} \cdot \lim_{x \to 0} \frac{2\sin^2(x)}{x^2} = 2 \cdot 2 = 4. \]

### Final Answer

Thus, the limit evaluates to:

\[ \boxed{4}. \]