Question

lim (1-x)tg $\pi x/2$
x->1

Answer 1

[tex] \displaystyle\lim_{x \to 1} (1-x)\cdot \frac{sin \frac{\pi x}{2} }{cos \frac{\pi x}{2}} =\\ =\lim_{x \to 1} sin \frac{\pi x}{2} \cdot \lim_{x \to 1} \frac{1-x }{cos \frac{\pi x}{2}}=\\ =\lim_{x \to 1} sin \frac{\pi x}{2} \cdot \lim_{x \to 1} \frac{1-x }{cos \frac{\pi x}{2}}=\\ =1 \cdot \lim_{x \to 1} \frac{1-x }{cos \frac{\pi x}{2}}=\\ = \lim_{x \to 1} \frac{1-x }{cos \frac{\pi x}{2}}=-Notam\ x-1=t\\ = \lim_{t \to 0} \frac{-t}{cos( \frac{\pi t}{2}+ \frac{\pi}{2} )} =\\ [/tex]
[tex]\displaystyle= \lim_{t \to 0} \frac{-t}{cos( \frac{\pi t}{2}+ \frac{\pi}{2} )} =\\ = \lim_{t \to 0} \frac{-t}{-sin \frac{\pi t}{2} } =\\ =\lim_{t \to 0} \frac{t}{sin \frac{\pi t}{2} } =\\ = \frac{1}{ \frac{\pi}{2} } =\\ =\frac{2}{\pi} [/tex]