Question

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

Hint:

For small angle limits involving sine or cosine, always expand using Taylor series up to the necessary order.

Answer 1

Step 1: Expand $\sin x$ using Taylor series.
\[ \sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots \] Step 2: Substitute expansion.
\[ x - \sin x = x - \left( x - \frac{x^3}{6} + \frac{x^5}{120} \right) = \frac{x^3}{6} - \frac{x^5}{120} + \cdots \] Step 3: Simplify expression.
\[ \frac{6(x - \sin x)}{x^3} = \frac{6\left(\frac{x^3}{6} - \frac{x^5}{120} + \cdots \right)}{x^3} \] \[ = \frac{x^3 - \frac{x^5}{20} + \cdots}{x^3} \] \[ = 1 - \frac{x^2}{20} + \cdots \] Step 4: Limit as $x$ tends to zero.
\[ \lim_{x \to 0} \frac{6(x - \sin x)}{x^3} = 1 \] \[ \boxed{\text{Final Answer = 1}} \]