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:
The problem presents an algebraic trigonometric limit evaluating to an indeterminate form of type $\frac{0}{0}$ as $x$ approaches $0$. We must evaluate the true limit.

Step 2: Key Formula or Approach:
We can solve this problem elegantly using two different approaches: 1. Trigonometric Factorization: Apply the identity $\cos C - \cos D = -2\sin\left(\frac{C+D}{2}\right)\sin\left(\frac{C-D}{2}\right)$ along with the standard limit $\lim_{\theta \to 0} \frac{\sin\theta}{\theta} = 1$. 2. L'Hôpital's Rule: Since it's a $\frac{0}{0}$ form, differentiate the numerator and denominator with respect to $x$ successively.

Step 3: Detailed Explanation:
Let's apply L'Hôpital's Rule for a crisp derivation. Differentiating the numerator and denominator once with respect to $x$: $$\text{Numerator derivative} = -m\sin(mx) - (-n\sin(nx)) = n\sin(nx) - m\sin(mx)$$ $$\text{Denominator derivative} = 2x$$ This gives: $$\lim_{x \to 0} \frac{n\sin(nx) - m\sin(mx)}{2x}$$ This is still in $\frac{0}{0}$ form. Differentiating a second time: $$\text{Numerator second derivative} = n^2\cos(nx) - m^2\cos(mx)$$ $$\text{Denominator second derivative} = 2$$ Now, substitute $x = 0$ safely into the limit expression: $$\lim_{x \to 0} \frac{n^2\cos(0) - m^2\cos(0)}{2} = \frac{n^2(1) - m^2(1)}{2} = \frac{n^2 - m^2}{2}$$

Step 4: Final Answer:
The limit value equals $\frac{n^2 - m^2}{2}$, which matches option (C).