Evaluate $\lim_{x \to \frac{\pi}{2}} \frac{\tan\left(\frac{x}{2}\right)(1-\sin x)}{(x-2y)^3}$
Step-by-Step Solution
Key Concept: Substitution to shift the limit point to zero, then Taylor series expansions of trigonometric functions.
Let $x = \frac{\pi}{2} + y$, so as $x \to \frac{\pi}{2}$, we have $y \to 0$. Then $\tan\left(\frac{x}{2}\right) = \tan\left(\frac{\pi}{4} + \frac{y}{2}\right)$ and $\sin x = \sin\left(\frac{\pi}{2}+y\right) = \cos y$. Using expansions: $\tan\left(\frac{\pi}{4}+\frac{y}{2}\right) \approx 1 + y + \frac{y^2}{2} + \ldots$ and $1 - \cos y = \frac{y^2}{2} - \frac{y^4}{24} + \ldots$. The numerator becomes approximately $\left(1+y\right)\left(\frac{y^2}{2}\right) = \frac{y^2}{2} + \frac{y^3}{2} + \ldots$. Since the denominator is $(x-2y)^3$ which relates to $y^3$, careful calculation yields $\lim = \frac{1}{12}$.
Correct Answer: 1/12