Question

100 баллов, ПРЕДЕЛЫ!!!
Решите пределы, не применяя производную!
ответы:
112: -1
107: 14
98: -2
106:-1/24
113: корень из 3
115: - корня из 2
131: 1

Answer 1

$115.\; \; \; \lim\limits _{x \to \pi /4}\frac{sin2x-cos2x-1}{cosx-sinx}=\lim\limits _{x \to \pi /4}\frac{2\, sinx\, cosx-(cos^2x-sin^2x)-(sin^2x+cos^2x)}{cosx-sinx}=\\\\=\lim\limits _{x \to \pi /4}\frac{2\, cosx\, (sinx-cosx)}{-(sinx-cosx)}=\lim\limits _{x \to \pi /4}\, (-2cosx)=-2\cdot cos\frac{\pi}{4}=-2\cdot \frac{\sqrt2}{2}=-\sqrt2$

$98.\; \; \; \lim\limits _{x \to 0}\frac{cos2x-1}{x^2}=\Big[\; 1-cos\alpha =2sin^2\frac{\alpha }{2}\; \Big]=\lim\limits _{x \to 0}\frac{-2sin^2x}{x^2}=\\\\=\Big[\; sinx\sim x\; ,\; esli\; x\to 0\; ;\; \; sin^2x=(sinx)^2\sim x^2\; \Big]=\lim\limits _{x \to 0}\frac{-2\cdot x^2}{x^2}=-2$

$107.\; \; \lim\limits _{x \to 0}\frac{sin7x}{\sqrt{1+x}-1}=\lim\limits _{x \to 0}\frac{7x\, \cdot \, (\sqrt{1+x}+1)}{(\sqrt{1+x}-1)(\sqrt{1+x}+1)}=\lim\limits _{x \to 0}\frac{7x\, \cdot \, (\sqrt{1+x}+1)}{(1+x)-1}=\\\\=\lim\limits_{x \to 0}\frac{7x\, \cdot \, (\sqrt{1+x}+1)}{x}=\lim\limits _{x \to 0}\, 7(\sqrt{1+x}+1)=7\cdot (\sqrt1+1)=7\cdot 2=14$

$106.\; \; \lim\limits _{x \to 0}\frac{\sqrt{9-x}-3}{sin4x}=\lim\limits _{x \to 0}\frac{(\sqrt{9-x}-3)(\sqrt{9-x}+3)}{4x\, \cdot \, (\sqrt{9-x}+3)}=\lim\limits _{x \to 0}\, \frac{(9-x)-9}{4x(\sqrt{9x-3}+3)}=\\\\=\lim\limits _{x \to 0}\, \frac{-1}{4\, (\sqrt{9-x}+3)}=-\frac{1}{4\cdot 6}=-\frac{1}{24}$

$131.\; \; \; \lim\limits _{x \to 0}\, (cosx)^{\frac{1}{sinx}}=\lim\limits _{x \to 0}\, \Big(1+(cosx-1)\Big)^{\frac{1}{cosx-1}\cdot \frac{cosx-1}{sinx}}=\\\\=\lim\limits _{x \to 0}\Big(\Big(1+(cosx-1)\Big)^{\frac{1}{cosx-1}}\Big)^{\frac{cosx-1}{sinx}}=e^{\lim\limits_{x \to 0}\frac{-(1-cosx)}{sinx}}=e^{\lim\limits_{x \to 0}\frac{-2sin^2x/2}{x}}=\\\\=e^{\lim\limits_{x \to 0}\frac{-2\cdot \frac{x^2}{4}}{x}}=e^{\lim\limits_{x \to 0}\frac{-x}{2}}=e^0=1$

$112.\; \; \lim\limits _{x \to \pi /2}\Big(x-\frac{\pi}{2}\Big)\cdodt tgx=\Big[\; t=x-\frac{\pi}{2}\; ,\; t\to 0\; \Big]=\lim\limits _{t \to 0}\; (\, t\cdot tg(t+\frac{\pi}{2})\, )=\\\\=\lim\limits _{t \to 0}\; \frac{t\cdot sin(t+\frac{\pi}{2})}{cos(t+\frac{\pi}{2})}=\lim\limits _{t \to 0}\; \frac{t\cdot sin(t+\frac{\pi}{2})}{-sint}=\lim\limits _{t \to 0}\; \frac{t\cdot sin(t+\frac{\pi}{2})}{-t}=\lim\limits _{t \to 0}\; (-sin\frac{\pi}{2})=-1$

$113.\; \; \lim\limits _{x \to \pi /6}\frac{1-2sinx}{\frac{\pi}{6}-x}=\lim\limits _{x \to \pi /6}\frac{2\, (\frac{1}{2}-sinx)}{\frac{\pi }{6} -x}=2\lim\limits _{x \to \po /6}\frac{sin\frac{\pi}{6}-sinx}{\frac{\pi}{6}-x}=\\\\=2\lim\limits _{x \to \pi /6}\frac{2\, sin(\frac{\pi}{12}-\frac{x}{2})\cdot cos(\frac{\pi}{12}+\frac{x}{2})}{\frac{\pi}{6}-x}=\Big[\; sina\sim a,\; a\to 0\; \Big]=4\lim\limits _{x \to \pi /6}\frac{(\frac{\pi}{12}-\frac{x}{2})\cdot cos(\frac{\pi}{12}+\frac{x}{2})}{\frac{\pi -6x}{6}}=$

$=4\lim\limits _{x \to \pi /6}\frac{\frac{\pi -6x}{12}\cdot cos\frac{\pi}{6}}{\frac{\pi -6x}{6}}=4\lim\limits_{x \to \\pi /6}\, \frac{(\pi -6x)\cdot \frac{\sqrt3}{2}}{12\cdot \frac{\pi -6x}{6}}=4\lim\limits _{x \to \pi /6}\frac{\sqrt3}{4}=\sqrt3$