Question

Evaluate the limit: $$ \lim_{x \to 0} \left( x - \sin x \right) \left( \frac{1}{x} \right) $$

• A. 0
• B. 1
• C. 2
• D. 3

Hint:

To evaluate limits involving trigonometric functions, you can use the Taylor series expansion near 0 to simplify the expression.

Answer 1

The Correct Option is A

Step 1: Use Taylor series expansion for \( \sin x \). 
The Taylor series expansion of \( \sin x \) around \( x = 0 \) is given by: \[ \sin x = x - \frac{x^3}{6} + O(x^5) \] So, \[ x - \sin x = x - \left( x - \frac{x^3}{6} + O(x^5) \right) = \frac{x^3}{6} + O(x^5) \] 
Step 2: Substitute into the limit expression. 
Substitute the result of \( x - \sin x \) into the original limit expression: \[ \lim_{x \to 0} \left( x - \sin x \right) \left( \frac{1}{x} \right) = \lim_{x \to 0} \left( \frac{x^3}{6} \cdot \frac{1}{x} \right) \] \[ = \lim_{x \to 0} \frac{x^2}{6} \] 
Step 3: Evaluate the limit. As \( x \to 0 \), \( \frac{x^2}{6} \to 0 \). 
Final Answer: The value of the limit is 0.