Evaluate the following :
`lim_(x to 0)(sin^(2)x)/(2x)`

Answer

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

### Step 1: Identify the form of the limit
When we substitute \( x = 0 \) into the expression, we get:
\[ \frac{\sin^2(0)}{2 \cdot 0} = \frac{0}{0} \]
This is an indeterminate form, so we can apply L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule
L'Hôpital's Rule states that if we have an indeterminate form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), we can differentiate the numerator and the denominator separately.

The numerator is \( \sin^2 x \) and the denominator is \( 2x \).

#### Differentiate the numerator:
Using the chain rule:
\[ \frac{d}{dx}(\sin^2 x) = 2 \sin x \cdot \cos x \]

#### Differentiate the denominator:
\[ \frac{d}{dx}(2x) = 2 \]

### Step 3: Rewrite the limit
Now we can rewrite the limit using the derivatives:
\[ \lim_{x \to 0} \frac{\sin^2 x}{2x} = \lim_{x \to 0} \frac{2 \sin x \cos x}{2} \]
The \( 2 \) in the numerator and denominator cancels out:
\[ = \lim_{x \to 0} \sin x \cos x \]

### Step 4: Evaluate the limit
Now we can substitute \( x = 0 \):
\[ \sin(0) \cdot \cos(0) = 0 \cdot 1 = 0 \]

### Final Answer
Thus, the limit is:
\[ \lim_{x \to 0} \frac{\sin^2 x}{2x} = 0 \]

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