Question

The value of \[ \lim_{x \to 0} \frac{6(x - \sin x)}{x^3} \] is ........ (in integer).

Hint:

When dealing with limits involving trigonometric functions, use the Taylor series expansions to approximate the function near the point of interest (here, \( x = 0 \)).

Answer 1

To evaluate the limit, we use the Taylor expansion of \( \sin x \) around \( x = 0 \): \[ \sin x = x - \frac{x^3}{6} + O(x^5) \] Now, substitute this into the expression: \[ 6(x - \sin x) = 6\left(x - \left(x - \frac{x^3}{6}\right)\right) = 6\left(\frac{x^3}{6}\right) = x^3 \] Thus, the expression becomes: \[ \frac{6(x - \sin x)}{x^3} = \frac{x^3}{x^3} = 1 \] Therefore, the value of the limit is 1.