`underset(x to 0)(lim) (1-cos x)/(x)` is :

Options

A. 0
B. 1
C. `(1)/(2)`
D. None of these

Answer

To find the limit as \( x \) approaches \( 0 \) for the expression \( \frac{1 - \cos x}{x} \), we can follow these steps:

### Step 1: Rewrite the expression
We start with the limit:
\[ \lim_{x \to 0} \frac{1 - \cos x}{x} \]

### Step 2: Multiply by the conjugate
To simplify the expression, we can multiply the numerator and the denominator by \( 1 + \cos x \):
\[ \lim_{x \to 0} \frac{(1 - \cos x)(1 + \cos x)}{x(1 + \cos x)} \]

### Step 3: Simplify the numerator
Using the identity \( 1 - \cos^2 x = \sin^2 x \), we can rewrite the numerator:
\[ \lim_{x \to 0} \frac{\sin^2 x}{x(1 + \cos x)} \]

### Step 4: Split the limit
We can separate the limit into two parts:
\[ \lim_{x \to 0} \left( \frac{\sin^2 x}{x} \cdot \frac{1}{1 + \cos x} \right) \]

### Step 5: Evaluate the first limit
Using the standard limit \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \), we can express \( \sin^2 x \) as:
\[ \lim_{x \to 0} \left( \frac{\sin x}{x} \cdot \sin x \right) = 1 \cdot \lim_{x \to 0} \sin x \]
As \( x \to 0 \), \( \sin x \to 0 \), so:
\[ \lim_{x \to 0} \sin^2 x = 0 \]

### Step 6: Evaluate the second limit
Now we evaluate the second part:
\[ \lim_{x \to 0} \frac{1}{1 + \cos x} \]
As \( x \to 0 \), \( \cos x \to 1 \), thus:
\[ \lim_{x \to 0} \frac{1}{1 + \cos x} = \frac{1}{2} \]

### Step 7: Combine the limits
Now we can combine the results:
\[ \lim_{x \to 0} \frac{\sin^2 x}{x} \cdot \lim_{x \to 0} \frac{1}{1 + \cos x} = 0 \cdot \frac{1}{2} = 0 \]

### Final Answer
Therefore, the limit is:
\[ \lim_{x \to 0} \frac{1 - \cos x}{x} = 0 \]

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