Question

Без Лопиталя как решать? ​

Answer 1

$\displaystyle \lim_{x \to 1}(1-x^{2})\,\text{tg}\, \frac{\pi x}{2} = [0 \cdot \infty] = \lim_{x \to 1}(1-x^{2})\dfrac{\sin \dfrac{\pi x}{2} }{\cos \dfrac{\pi x}{2} } =$

$\displaystyle = \underset{1}{\underbrace{\lim_{x \to 1} \sin \dfrac{\pi x}{2}}} \cdot \lim_{x \to 1}\frac{1 - x^{2}}{\cos \dfrac{\pi x}{2} } = \lim_{x \to 1}\frac{1 - x^{2}}{\cos \dfrac{\pi x}{2} } = \left[\frac{0}{0} \right] =$

$\displaystyle = \left|\begin{array}{ccc}x = 1 - t\\t = 1 - x\\t \to 0\end{array}\right| = \lim_{t \to 0} \frac{1 - (1 - t)^{2}}{\cos \dfrac{\pi(1-t)}{2} } = \lim_{t \to 0} \frac{t(2-t)}{\cos\left(\dfrac{\pi}{2} - \dfrac{\pi t}{2} \right)} =$

$= \displaystyle \lim_{t \to 0}\frac{t(2-t)}{\sin \dfrac{\pi t}{2} } = \left | {{\sin \dfrac{\pi t}{2} \sim \dfrac{\pi t}{2} } \atop {t \to 0}} \right | = \lim _{t \to 0} \frac{t(2-t)}{\dfrac{\pi t}{2} } = \lim_{t \to 0} \frac{4 - 2t}{\pi} =\frac{4}{\pi}$

Ответ: $\dfrac{4}{\pi}$