Пределы найти:1) lim x -> 02) lim x-> 03) lim x-> 0
Второй предел слева или справа нужно найти?
Метод замены бесконечно малых величин эквивалентными бесконечно малыми.
( Если , то бесконечно малая. )
Решение во вложении
, то 
бесконечно малая. )![1)\; \; \lim\limits _{x \to 0}\frac{1-sinx-cosx}{sin(\sqrt2x)}=\lim\limits _{x \to 0}\frac{1-cosx}{sin(\sqrt2x)}-\lim\limits _{x \to 0}\frac{sinx}{sin(\sqrt2x)}=\\\\=\Big [\, (1-cos\alpha )\sim \frac{\alpha ^2}{2}\; ,\; \; sin\alpha \sim \alpha \; ,\; \; \alpha \to 0\; \Big ]=\\\\=\lim\limits _{x \to 0}\frac{\frac{x^2}{2}}{\sqrt2x}-\lim\limits _{x \to 0}\frac{x}{\sqrt2x}=\lim\limits_{x \to 0}\frac{x}{2\sqrt2}-\lim\limits _{x \to 0}\frac{1}{\sqrt2}=\frac{0}{2\sqrt2}-\frac{1}{\sqrt2}=-\frac{1}{\sqrt2}](https://tex.z-dn.net/?f=1%29%5C%3B%20%5C%3B%20%5Clim%5Climits%20_%7Bx%20%5Cto%200%7D%5Cfrac%7B1-sinx-cosx%7D%7Bsin%28%5Csqrt2x%29%7D%3D%5Clim%5Climits%20_%7Bx%20%5Cto%200%7D%5Cfrac%7B1-cosx%7D%7Bsin%28%5Csqrt2x%29%7D-%5Clim%5Climits%20_%7Bx%20%5Cto%200%7D%5Cfrac%7Bsinx%7D%7Bsin%28%5Csqrt2x%29%7D%3D%5C%5C%5C%5C%3D%5CBig%20%5B%5C%2C%20%281-cos%5Calpha%20%29%5Csim%20%5Cfrac%7B%5Calpha%20%5E2%7D%7B2%7D%5C%3B%20%2C%5C%3B%20%5C%3B%20sin%5Calpha%20%5Csim%20%5Calpha%20%5C%3B%20%2C%5C%3B%20%5C%3B%20%5Calpha%20%5Cto%200%5C%3B%20%5CBig%20%5D%3D%5C%5C%5C%5C%3D%5Clim%5Climits%20_%7Bx%20%5Cto%200%7D%5Cfrac%7B%5Cfrac%7Bx%5E2%7D%7B2%7D%7D%7B%5Csqrt2x%7D-%5Clim%5Climits%20_%7Bx%20%5Cto%200%7D%5Cfrac%7Bx%7D%7B%5Csqrt2x%7D%3D%5Clim%5Climits_%7Bx%20%5Cto%200%7D%5Cfrac%7Bx%7D%7B2%5Csqrt2%7D-%5Clim%5Climits%20_%7Bx%20%5Cto%200%7D%5Cfrac%7B1%7D%7B%5Csqrt2%7D%3D%5Cfrac%7B0%7D%7B2%5Csqrt2%7D-%5Cfrac%7B1%7D%7B%5Csqrt2%7D%3D-%5Cfrac%7B1%7D%7B%5Csqrt2%7D "1)\; \; \lim\limits _{x \to 0}\frac{1-sinx-cosx}{sin(\sqrt2x)}=\lim\limits _{x \to 0}\frac{1-cosx}{sin(\sqrt2x)}-\lim\limits _{x \to 0}\frac{sinx}{sin(\sqrt2x)}=\\\\=\Big [\, (1-cos\alpha )\sim \frac{\alpha ^2}{2}\; ,\; \; sin\alpha \sim \alpha \; ,\; \; \alpha \to 0\; \Big ]=\\\\=\lim\limits _{x \to 0}\frac{\frac{x^2}{2}}{\sqrt2x}-\lim\limits _{x \to 0}\frac{x}{\sqrt2x}=\lim\limits_{x \to 0}\frac{x}{2\sqrt2}-\lim\limits _{x \to 0}\frac{1}{\sqrt2}=\frac{0}{2\sqrt2}-\frac{1}{\sqrt2}=-\frac{1}{\sqrt2}")
![2)\; \; \lim\limits _{x \to 0}\frac{ln(1+x)}{x^2}=\Big [\; ln(1+\alpha )\sim \alpha \; ,esli\; \; \alpha \to 0\; \Big ]=\lim\limits _{x \to 0}\frac{x}{x^2}=\\\\=\lim\limits _{x \to 0}\frac{1}{x}=\Big [\; \frac{1}{0}\; \Big ]=\infty \\\\3)\; \; \lim\limits _{x \to 0}\frac{tgx}{2x}=\Big [\; tg\alpha \sim \alpha \; ,\; esli\; \alpha \to 0\; \Big ]=\lim\limits _{x \to 0}\frac{x}{2x}=\frac{1}{2}](https://tex.z-dn.net/?f=2%29%5C%3B%20%5C%3B%20%5Clim%5Climits%20_%7Bx%20%5Cto%200%7D%5Cfrac%7Bln%281%2Bx%29%7D%7Bx%5E2%7D%3D%5CBig%20%5B%5C%3B%20ln%281%2B%5Calpha%20%29%5Csim%20%5Calpha%20%5C%3B%20%2Cesli%5C%3B%20%5C%3B%20%5Calpha%20%5Cto%200%5C%3B%20%5CBig%20%5D%3D%5Clim%5Climits%20_%7Bx%20%5Cto%200%7D%5Cfrac%7Bx%7D%7Bx%5E2%7D%3D%5C%5C%5C%5C%3D%5Clim%5Climits%20_%7Bx%20%5Cto%200%7D%5Cfrac%7B1%7D%7Bx%7D%3D%5CBig%20%5B%5C%3B%20%20%5Cfrac%7B1%7D%7B0%7D%5C%3B%20%5CBig%20%5D%3D%5Cinfty%20%5C%5C%5C%5C3%29%5C%3B%20%5C%3B%20%5Clim%5Climits%20_%7Bx%20%5Cto%200%7D%5Cfrac%7Btgx%7D%7B2x%7D%3D%5CBig%20%5B%5C%3B%20tg%5Calpha%20%5Csim%20%5Calpha%20%5C%3B%20%2C%5C%3B%20esli%5C%3B%20%5Calpha%20%5Cto%200%5C%3B%20%5CBig%20%5D%3D%5Clim%5Climits%20_%7Bx%20%5Cto%200%7D%5Cfrac%7Bx%7D%7B2x%7D%3D%5Cfrac%7B1%7D%7B2%7D "2)\; \; \lim\limits _{x \to 0}\frac{ln(1+x)}{x^2}=\Big [\; ln(1+\alpha )\sim \alpha \; ,esli\; \; \alpha \to 0\; \Big ]=\lim\limits _{x \to 0}\frac{x}{x^2}=\\\\=\lim\limits _{x \to 0}\frac{1}{x}=\Big [\; \frac{1}{0}\; \Big ]=\infty \\\\3)\; \; \lim\limits _{x \to 0}\frac{tgx}{2x}=\Big [\; tg\alpha \sim \alpha \; ,\; esli\; \alpha \to 0\; \Big ]=\lim\limits _{x \to 0}\frac{x}{2x}=\frac{1}{2}")