Question

The value of \(\lim_{x \to 0^+} \left(\frac{1}{x} - \frac{1}{\sin x}\right)\) is:

• A. 1
• B. Infinite
• C. -1
• D. 0

Hint:

Exam Tip:
For limits of the form \(\infty - \infty\):

• Combine into a single fraction.
• Use Taylor series expansions.
• L'Hôpital's rule can also be used.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
We need to evaluate the limit of the difference of two terms that both tend to infinity as \(x \to 0^+\). We need to simplify and find the finite value.

Step 2: Key Formula or Approach:
Use the Taylor series expansion of \(\sin x\): \(\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots\).

Step 3: Detailed Explanation:
\[ \frac{1}{x} - \frac{1}{\sin x} = \frac{\sin x - x}{x \sin x} \] Use the expansion \(\sin x = x - \frac{x^3}{6} + \mathcal{O}(x^5)\): \[ \sin x - x = -\frac{x^3}{6} + \mathcal{O}(x^5) \] \[ x \sin x = x\left(x - \frac{x^3}{6} + \cdots\right) = x^2 - \frac{x^4}{6} + \cdots \] So, \[ \frac{\sin x - x}{x \sin x} = \frac{-\frac{x^3}{6} + \mathcal{O}(x^5)}{x^2 - \frac{x^4}{6} + \cdots} = \frac{-\frac{x}{6} + \mathcal{O}(x^3)}{1 - \frac{x^2}{6} + \cdots} \to 0 \] Thus, the limit is 0.

Step 4: Final Answer:
Therefore, option (D) is correct.