$\lim_{x \to 0} \frac{1 + \sin x - \cos x - \sin(x) \cdot x}{x \sin^2 x}$
Step-by-Step Solution
Key Concept: Use Taylor series expansions around $x=0$ or apply L'Hôpital's rule repeatedly for $\frac{0}{0}$ indeterminate forms.
For $\lim_{x \to 0} \frac{1 + \sin x - \cos x - \sin(x) \cdot x}{x \sin^2 x}$, using Taylor series: $\sin x = x - \frac{x^3}{6} + ...$, $\cos x = 1 - \frac{x^2}{2} + ...$. The numerator becomes $1 + x - \frac{x^3}{6} - (1 - \frac{x^2}{2}) - x(x - \frac{x^3}{6}) = \frac{x^2}{2} - \frac{x^2}{1} + O(x^3) = \frac{x^2}{2} + O(x^3)$. The denominator is $x \cdot x^2 = x^3 + O(x^5)$, but careful expansion gives the limit as $\frac{1}{2}$ using L'Hôpital's rule thrice.
Correct Answer: 1