What is the value of `lim_(xto0)(2(1-cosx))/(x^(2))`?

Options

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

Answer

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

### Step 1: Rewrite the expression using the trigonometric identity
We know that \( 1 - \cos x = 2 \sin^2\left(\frac{x}{2}\right) \). Therefore, we can rewrite the limit as:
\[ \lim_{x \to 0} \frac{2(1 - \cos x)}{x^2} = \lim_{x \to 0} \frac{2 \cdot 2 \sin^2\left(\frac{x}{2}\right)}{x^2} = \lim_{x \to 0} \frac{4 \sin^2\left(\frac{x}{2}\right)}{x^2} \]

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

### Step 3: Apply the limit property
Using the limit property \( \lim_{u \to 0} \frac{\sin u}{u} = 1 \), we can substitute \( u = \frac{x}{2} \):
\[ \lim_{x \to 0} \frac{\sin^2\left(\frac{x}{2}\right)}{\left(\frac{x}{2}\right)^2} = \left( \lim_{u \to 0} \frac{\sin u}{u} \right)^2 = 1^2 = 1 \]

### Step 4: Combine the results
Now, we can combine the results:
\[ \lim_{x \to 0} \frac{2(1 - \cos x)}{x^2} = 4 \cdot 1 = 2 \]

### Final Answer
Thus, the value of the limit is:
\[ \lim_{x \to 0} \frac{2(1 - \cos x)}{x^2} = 2 \]