Question

\[ \lim_{x \to 0} \left( \frac{1}{\sin x} - \frac{1}{x} \right) = \underline{\hspace{2cm}} \text{ (round off to nearest integer).} \]

Hint:

Use series expansion of \(\sin x\) for limits involving differences like \(x - \sin x\).

Answer 1

Correct Answer: 0

Rewrite the expression:
\[ \frac{1}{\sin x} - \frac{1}{x} = \frac{x - \sin x}{x \sin x}. \] Using the standard limit expansion: \[ \sin x = x - \frac{x^3}{6} + O(x^5), \] \[ x - \sin x = \frac{x^3}{6} + O(x^5). \] Thus, \[ \frac{x - \sin x}{x \sin x} \approx \frac{\frac{x^3}{6}}{x^2} = \frac{x}{6} \to 0. \] Hence the limit is approximately \(0\). Rounded to nearest integer, the answer is \(0\).