GATE 2023 BT – Question 31

Correct answer: 6

Explanation

The limit is of the type $0/0$ at $x=0$, so use the series for cosine:
$$\cos ax=1-\frac{a^2x^2}{2}+\frac{a^4x^4}{24}-\cdots$$

**Numerator:**
$$\cos2x-\cos4x=\left(1-2x^2+\cdots\right)-\left(1-8x^2+\cdots\right)=6x^2+O(x^4).$$

**Limit:**
$$\lim_{x\to0}\frac{6x^2+O(x^4)}{x^2}=\mathbf{6}.$$

(Check by L'Hôpital: twice differentiating gives $(-4\cos2x+16\cos4x)/2\to(−4+16)/2=6$.)