Question

The value of \[ \lim_{x \to 0} \frac{1 - \cos x}{x^2} \] is

• A. \( \frac{1}{4} \)
• B. \( \frac{1}{3} \)
• C. \( \frac{1}{2} \)
• D. 1

Hint:

When evaluating limits, use Taylor series expansions for standard functions like \( \cos x \) around \( x = 0 \) to simplify the expression.

Answer 1

The Correct Option is C

We are asked to find the value of the limit: \[ \lim_{x \to 0} \frac{1 - \cos x}{x^2}. \] We can use the Taylor series expansion of \( \cos x \) around \( x = 0 \) to solve this. The Taylor series for \( \cos x \) is: \[ \cos x = 1 - \frac{x^2}{2} + O(x^4). \] Now substitute this expansion into the given expression: \[ 1 - \cos x = 1 - \left( 1 - \frac{x^2}{2} + O(x^4) \right) = \frac{x^2}{2} + O(x^4). \] Thus, the limit becomes: \[ \lim_{x \to 0} \frac{\frac{x^2}{2} + O(x^4)}{x^2} = \lim_{x \to 0} \frac{x^2}{2x^2} = \frac{1}{2}. \] Therefore, the value of the limit is \( \frac{1}{2} \), corresponding to Option (C).
Final Answer: (C) \( \frac{1}{2} \)