Question

Вычислить пределы (использовать максимум эквивалентные преобразования):

$1.\ \lim_{x \to 0} \frac{tg\ x-sin\ x}{x(1-cos\ 2x)} \\\\2.\ \lim_{x \to \pi/2} \frac{1-sin\ x}{(\pi/2-x)^2}$

Answer 1

$\displaystyle~1)\lim_{x \to 0}\frac{{\rm tg}\, x-\sin x}{x(1-\cos 2x)}=\lim_{x \to 0}\frac{\sin x\left(\frac{1}{\cos x}-1\right)}{x(1-1+2\sin^2x)}=\lim_{x \to 0}\frac{x\cdot \left(\frac{1}{\cos x}-1\right)}{x\cdot 2\sin^2x}=\\ \\ \\ =\lim_{x \to 0}\frac{1-\cos x}{2\cos x\sin^2x}=\lim_{x \to 0}\frac{1-\cos x}{2\cos x(1-\cos x)(1+\cos x)}=\\ \\ \\ =\lim_{x \to 0}\frac{1}{2\cos x(1+\cos x)}=\dfrac{1}{2\cdot 1\cdot (1+1)}=\dfrac{1}{4}$

$\displaystyle 2)~ \lim_{x \to \frac{\pi}{2}}\frac{1-\sin x}{(\frac{\pi}{2}-x)^2}=\lim_{x \to \frac{\pi}{2}}\frac{1-\cos \left(\frac{\pi}{2}-x\right)}{\left(\frac{\pi}{2}-x\right)^2}=\\ \\ \\ =\lim_{x \to \frac{\pi}{2}}\frac{\Big(1-\cos \left(\frac{\pi}{2}-x\right)\Big)\Big(1+\cos\left(\frac{\pi}{2}-x\right)\Big)}{\Big(1+\cos\left(\frac{\pi}{2}-x\right)\Big)\left(\frac{\pi}{2}-x\right)^2}=\lim_{x \to \frac{\pi}{2}}\frac{\sin^2\left(\frac{\pi}{2}-x\right)}{\Big(1+\cos\left(\frac{\pi}{2}-x\right)\Big)\left(\frac{\pi}{2}-x\right)^2}$

$=\displaystyle \lim_{x \to \frac{\pi}{2}}\frac{\left(\frac{\pi}{2}-x\right)^2}{\Big(1+\cos\left(\frac{\pi}{2}-x\right)\Big)\left(\frac{\pi}{2}-x\right)^2}=\dfrac{1}{1+1}=\dfrac{1}{2}$