GATE 2021 AE – Question 27

Correct answer: 0

Explanation

**Series expansion.** Near $x=0$,
$$\sin x=x-\frac{x^3}{6}+\cdots\;\Rightarrow\;\frac{1}{\sin x}=\frac1x\left(1-\frac{x^2}{6}+\cdots\right)^{-1}=\frac1x+\frac{x}{6}+\cdots$$

**Difference:**
$$\frac1{\sin x}-\frac1x=\frac x6+O(x^3)\;\longrightarrow\;\mathbf{0}\quad(x\to0).$$

(Alternatively, combine the fractions: $\dfrac{x-\sin x}{x\sin x}$. For small $x$ this is $\dfrac{x^3/6}{x^2}=\dfrac x6\to0$.)