`lim_(x to 0)(1- cos x)/x^(2)`

Options

A. 0
B. `1//4`
C. `1//2`
D. 1

Answer

To solve the limit \( \lim_{x \to 0} \frac{1 - \cos x}{x^2} \), we can follow these steps:

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

### Step 2: Simplify the limit
Next, we can simplify the limit:
\[ \lim_{x \to 0} \frac{2 \sin^2\left(\frac{x}{2}\right)}{x^2} = \lim_{x \to 0} \frac{2 \sin^2\left(\frac{x}{2}\right)}{\left(\frac{x}{2}\right)^2} \cdot \frac{1}{4} \]
This can be rewritten as:
\[ = \lim_{x \to 0} \frac{2 \sin^2\left(\frac{x}{2}\right)}{\left(\frac{x}{2}\right)^2} \cdot \frac{1}{4} = \frac{1}{2} \lim_{x \to 0} \frac{\sin^2\left(\frac{x}{2}\right)}{\left(\frac{x}{2}\right)^2} \]

### Step 3: Apply the limit property
Using the limit property \( \lim_{u \to 0} \frac{\sin u}{u} = 1 \), we can set \( u = \frac{x}{2} \). As \( x \to 0 \), \( u \to 0 \) as well:
\[ \lim_{x \to 0} \frac{\sin^2\left(\frac{x}{2}\right)}{\left(\frac{x}{2}\right)^2} = 1 \]
Thus, we have:
\[ \lim_{x \to 0} \frac{2 \sin^2\left(\frac{x}{2}\right)}{x^2} = \frac{1}{2} \cdot 1 = \frac{1}{2} \]

### Final Result
Therefore, the limit is:
\[ \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2} \]

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