Evaluate the following limits:
`lim_(xrarr0)((1-cos2x))/(sin^(2)2x)`

Answer

To evaluate the limit

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

we can follow these steps:

### Step 1: Rewrite the limit using trigonometric identities

We know that \(1 - \cos(2x) = 2\sin^2(x)\). Therefore, we can rewrite the limit as:

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

### Step 2: Simplify \(\sin^2(2x)\)

Using the identity \(\sin(2x) = 2\sin(x)\cos(x)\), we can express \(\sin^2(2x)\) as:

\[ \sin^2(2x) = (2\sin(x)\cos(x))^2 = 4\sin^2(x)\cos^2(x). \]

### Step 3: Substitute back into the limit

Now substitute \(\sin^2(2x)\) back into the limit:

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

### Step 4: Cancel out \(\sin^2(x)\)

Assuming \(x \neq 0\), we can cancel \(\sin^2(x)\) from the numerator and denominator:

\[ \lim_{x \to 0} \frac{2}{4\cos^2(x)} = \lim_{x \to 0} \frac{1}{2\cos^2(x)}. \]

### Step 5: Evaluate the limit

As \(x\) approaches 0, \(\cos^2(x)\) approaches \(\cos^2(0) = 1\). Therefore, we have:

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

### Final Answer

Thus, the limit is

\[ \frac{1}{2}. \]

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