Question

$$ \lim_{x \to 0} \frac{1 - \cos x}{x^2} = $$

• A. $\frac{1}{3}$
• B. $\frac{1}{2}$
• C. 3
• D. 2

Hint:

Memorize $\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$ as a standard result. It appears frequently as a sub-component in more complex calculus problems.

Answer 1

The Correct Option is B

To solve the limit problem \(\lim_{x \to 0} \frac{1 - \cos x}{x^2}\), we will use the Taylor series expansion for \(\cos x\).

1. The Taylor series expansion for \(\cos x\) around x = 0 is:
   \(\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \cdots\)
   For small values of x, higher-order terms become negligible.
2. Therefore, the expression \(1 - \cos x\) can be approximated by:
   \(1 - \cos x \approx 1 - \left(1 - \frac{x^2}{2}\right) = \frac{x^2}{2}\)
3. Substituting this approximation into the limit, we have:
   \(\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \lim_{x \to 0} \frac{\frac{x^2}{2}}{x^2} \)
4. Simplifying the expression inside the limit gives:
   \(\frac{x^2/2}{x^2} = \frac{1}{2}\)
5. Thus, the limit evaluates to:
   \(\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}\)

Therefore, the correct answer is \(\frac{1}{2}\).