Помогите！ Решить по правилу Лопиталя.
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![\lim\limits_{x\to\frac{\pi}{2a}}\frac{1-\sin{ax}}{(2ax-\pi)^2} =\lim\limits_{x\to\frac{\pi}{2a}}\frac{(1-\sin{ax})'}{[(2ax-\pi)^2]'}=\lim\limits_{x\to\frac{\pi}{2a}}\frac{-a\cos{ax}}{2\cdot2(2ax-\pi)} =-\frac{1}{4} \lim\limits_{x\to\frac{\pi}{2a}}\frac{(\cos{ax})'}{(2ax-\pi)'} =\\=-\frac{1}{4} \lim\limits_{x\to\frac{\pi}{2a}}\frac{-a\sin{ax}}{2a}=\frac{1}{8} \lim\limits_{x\to\frac{\pi}{2a}}\sin{ax}=\frac{1}{8} \lim\limits_{x\to\frac{\pi}{2a}}\sin\frac{\pi}{2} =\frac{1}{8}](https://tex.z-dn.net/?f=%5Clim%5Climits_%7Bx%5Cto%5Cfrac%7B%5Cpi%7D%7B2a%7D%7D%5Cfrac%7B1-%5Csin%7Bax%7D%7D%7B%282ax-%5Cpi%29%5E2%7D%20%3D%5Clim%5Climits_%7Bx%5Cto%5Cfrac%7B%5Cpi%7D%7B2a%7D%7D%5Cfrac%7B%281-%5Csin%7Bax%7D%29%27%7D%7B%5B%282ax-%5Cpi%29%5E2%5D%27%7D%3D%5Clim%5Climits_%7Bx%5Cto%5Cfrac%7B%5Cpi%7D%7B2a%7D%7D%5Cfrac%7B-a%5Ccos%7Bax%7D%7D%7B2%5Ccdot2%282ax-%5Cpi%29%7D%20%3D-%5Cfrac%7B1%7D%7B4%7D%20%5Clim%5Climits_%7Bx%5Cto%5Cfrac%7B%5Cpi%7D%7B2a%7D%7D%5Cfrac%7B%28%5Ccos%7Bax%7D%29%27%7D%7B%282ax-%5Cpi%29%27%7D%20%3D%5C%5C%3D-%5Cfrac%7B1%7D%7B4%7D%20%5Clim%5Climits_%7Bx%5Cto%5Cfrac%7B%5Cpi%7D%7B2a%7D%7D%5Cfrac%7B-a%5Csin%7Bax%7D%7D%7B2a%7D%3D%5Cfrac%7B1%7D%7B8%7D%20%5Clim%5Climits_%7Bx%5Cto%5Cfrac%7B%5Cpi%7D%7B2a%7D%7D%5Csin%7Bax%7D%3D%5Cfrac%7B1%7D%7B8%7D%20%5Clim%5Climits_%7Bx%5Cto%5Cfrac%7B%5Cpi%7D%7B2a%7D%7D%5Csin%5Cfrac%7B%5Cpi%7D%7B2%7D%20%3D%5Cfrac%7B1%7D%7B8%7D "\lim\limits_{x\to\frac{\pi}{2a}}\frac{1-\sin{ax}}{(2ax-\pi)^2} =\lim\limits_{x\to\frac{\pi}{2a}}\frac{(1-\sin{ax})'}{[(2ax-\pi)^2]'}=\lim\limits_{x\to\frac{\pi}{2a}}\frac{-a\cos{ax}}{2\cdot2(2ax-\pi)} =-\frac{1}{4} \lim\limits_{x\to\frac{\pi}{2a}}\frac{(\cos{ax})'}{(2ax-\pi)'} =\\=-\frac{1}{4} \lim\limits_{x\to\frac{\pi}{2a}}\frac{-a\sin{ax}}{2a}=\frac{1}{8} \lim\limits_{x\to\frac{\pi}{2a}}\sin{ax}=\frac{1}{8} \lim\limits_{x\to\frac{\pi}{2a}}\sin\frac{\pi}{2} =\frac{1}{8}")