lim(1+sinx)^1/x при x->0
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Ответил sweetburning
0
Применил таблицу эквивалентности
![lim_{x to 0} (1+sinx)^ frac{1}{x}= lim_{x to 0} (1+x)^ frac{1}{x}= \
= lim_{t to infty}(1+ frac{1}{t})^t=e](https://tex.z-dn.net/?f=+lim_%7Bx+to+0%7D+%281%2Bsinx%29%5E+frac%7B1%7D%7Bx%7D%3D+lim_%7Bx+to+0%7D+%281%2Bx%29%5E+frac%7B1%7D%7Bx%7D%3D+%5C+%0A%3D+lim_%7Bt+to+infty%7D%281%2B+frac%7B1%7D%7Bt%7D%29%5Et%3De++++++ "lim_{x to 0} (1+sinx)^ frac{1}{x}= lim_{x to 0} (1+x)^ frac{1}{x}= \
= lim_{t to infty}(1+ frac{1}{t})^t=e")
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