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

Options

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

Answer

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

### Step 1: Rewrite the expression using a trigonometric identity
We know from trigonometric identities that:
\[ 1 - \cos^2 x = \sin^2 x \]
This allows us to express \(1 - \cos x\) in terms of \(\sin x\):
\[ 1 - \cos x = \sin^2 x / (1 + \cos x) \]
Thus, we can rewrite the limit as:
\[ \lim_{x \to 0} \frac{1 - \cos x}{x} = \lim_{x \to 0} \frac{\sin^2 x}{(1 + \cos x)x} \]

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

### Step 3: Evaluate the first limit
We know from the standard limit result that:
\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \]

### Step 4: Evaluate the second limit
Now we need to evaluate:
\[ \lim_{x \to 0} \frac{\sin x}{1 + \cos x} \]
At \(x = 0\):
- \(\sin 0 = 0\)
- \(\cos 0 = 1\), so \(1 + \cos 0 = 2\)

Thus,
\[ \lim_{x \to 0} \frac{\sin x}{1 + \cos x} = \frac{0}{2} = 0 \]

### Step 5: Combine the results
Now we can combine the results from the two limits:
\[ \lim_{x \to 0} \frac{1 - \cos x}{x} = \left(\lim_{x \to 0} \frac{\sin x}{x}\right) \cdot \left(\lim_{x \to 0} \frac{\sin x}{1 + \cos x}\right) = 1 \cdot 0 = 0 \]

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

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