Calculus

### **Step-by-Step Solution & Mathematical Derivation (GATE BT 2023 Q.31)** **1. Governing Concept & Definition**: - **Topic**: Engineering Mathematics & Limits of Trigonometric Functions - Evaluating the indeterminate limit of the form $\left[\frac{0}{0}\right]$: $\lim_{x \to 0} \frac{\cos 2x - \cos 4x}{x^2}$ - Using the standard Taylor series expansion: $\cos(u) = 1 - \frac{u^2}{2} + O(u^4)$ **2. Detailed Step-by-Step Calculation**: - **Taylor expansions around $x = 0$**: $\cos(2x) = 1 - \frac{(2x)^2}{2} + O(x^4) = 1 - 2x^2 + O(x^4)$ $\cos(4x) = 1 - \frac{(4x)^2}{2} + O(x^4) = 1 - 8x^2 + O(x^4)$ - **Numerator**: $\cos(2x) - \cos(4x) = (1 - 2x^2) - (1 - 8x^2) = 6x^2 + O(x^4)$ - **Evaluating the limit**: $\lim_{x \to 0} \frac{6x^2 + O(x^4)}{x^2} = \lim_{x \to 0} \left(6 + O(x^2)\right) = 6$ - *(Alternative via L'Hôpital's Rule)*: $\lim_{x \to 0} \frac{-2\sin 2x + 4\sin 4x}{2x} = \lim_{x \to 0} \frac{-4\cos 2x + 16\cos 4x}{2} = \frac{-4(1) + 16(1)}{2} = \frac{12}{2} = 6$ **3. Final Answer**: - **Calculated Value**: **6** - **Official GATE Accepted Range**: **6 to 6**