Question

The value of $\lim_{x \to 0} \frac{\cos(mx) - \cos(nx)}{x^2}$ is

• A. $\frac{m^2 - n^2}{2}$
• B. $m^2 - n^2$
• C. $\frac{n^2 - m^2}{2}$
• D. $n^2 - m^2$

Hint:

Remember the expansion shortcut for cosine near zero: $\cos(\theta) \approx 1 - \frac{\theta^2}{2}$.
Substituting this approximation into our limit yields: $$\frac{\left(1 - \frac{(mx)^2}{2}\right) - \left(1 - \frac{(nx)^2}{2}\right)}{x^2} = \frac{-\frac{m^2x^2}{2} + \frac{n^2x^2}{2}}{x^2} = \frac{n^2 - m^2}{2}$$ This structure appears frequently, making it a valuable shortcut to memorize!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
Evaluate the limit lim_{x→0} (cos mx - cos nx)/x², which is of the 0/0 indeterminate form.

Step 2: Key Formula or Approach:
Apply L'Hôpital's Rule twice, differentiating numerator and denominator successively until the indeterminate form resolves.

Step 3: Detailed Explanation:
First differentiation: lim_{x→0} [n sin(nx) - m sin(mx)]/(2x) → still 0/0. Second differentiation: lim_{x→0} [n² cos(nx) - m² cos(mx)]/2 = [n²(1) - m²(1)]/2 = (n² - m²)/2.

Step 4: Final Answer:
The limit is (n² - m²)/2, option (C).