What is `lim_(xto0) (1-cosx)/(x)` equal to ?

Options

A. 0
B. `(1)/(2)`
C. 1
D. 2

Explanation

To find the limit \( \lim_{x \to 0} \frac{1 - \cos x}{x} \), we can follow these steps:

### Step 1: Rewrite the expression using a trigonometric identity
We know that \( 1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right) \). Therefore, we can rewrite the limit as:
\[ \lim_{x \to 0} \frac{1 - \cos x}{x} = \lim_{x \to 0} \frac{2 \sin^2\left(\frac{x}{2}\right)}{x} \]

### Step 2: Simplify the limit
Next, we can express \( x \) in terms of \( \frac{x}{2} \):
\[ \lim_{x \to 0} \frac{2 \sin^2\left(\frac{x}{2}\right)}{x} = \lim_{x \to 0} \frac{2 \sin^2\left(\frac{x}{2}\right)}{\frac{x}{2}} \cdot \frac{1}{2} \]
This gives us:
\[ = \lim_{x \to 0} \frac{2 \sin^2\left(\frac{x}{2}\right)}{\frac{x}{2}} \cdot \frac{1}{2} \]

### Step 3: Apply the limit property
We know from the standard limit that \( \lim_{u \to 0} \frac{\sin u}{u} = 1 \). Let \( u = \frac{x}{2} \), then as \( x \to 0 \), \( u \to 0 \) as well. Therefore:
\[ \lim_{x \to 0} \frac{\sin\left(\frac{x}{2}\right)}{\frac{x}{2}} = 1 \]

### Step 4: Substitute back into the limit
Now we can substitute this back into our limit:
\[ = 2 \cdot 1 \cdot \frac{1}{2} = 1 \]

### Step 5: Final evaluation
Thus, we conclude that:
\[ \lim_{x \to 0} \frac{1 - \cos x}{x} = 0 \]

### Final Answer
The limit is \( 0 \).

---