Помогите, нужно подробное решение
Приложения:
Ответы на вопрос
Ответил pavlikleon
0
избавимся от степени
![= lim_{n to infty} e^{ln ( frac{x+3}{x-2} )^{2x+1}}= \ \
=e^{ lim_{n to infty} ln ( frac{x+3}{x-2} )^{2x+1}}= \ \
=e^{ lim_{n to infty} ((2x+1)ln ( frac{x+3}{x-2} ))}= \ \](https://tex.z-dn.net/?f=%3D+lim_%7Bn+to+infty%7D+e%5E%7Bln+%28+frac%7Bx%2B3%7D%7Bx-2%7D++%29%5E%7B2x%2B1%7D%7D%3D+%5C++%5C+%0A%3De%5E%7B+lim_%7Bn+to+infty%7D+ln+%28+frac%7Bx%2B3%7D%7Bx-2%7D++%29%5E%7B2x%2B1%7D%7D%3D+%5C++%5C+%0A%3De%5E%7B+lim_%7Bn+to+infty%7D+%28%282x%2B1%29ln+%28+frac%7Bx%2B3%7D%7Bx-2%7D++%29%29%7D%3D+%5C++%5C+ "= lim_{n to infty} e^{ln ( frac{x+3}{x-2} )^{2x+1}}= \ \
=e^{ lim_{n to infty} ln ( frac{x+3}{x-2} )^{2x+1}}= \ \
=e^{ lim_{n to infty} ((2x+1)ln ( frac{x+3}{x-2} ))}= \ \")
под пределом получили неопределенность 0*∞ , преобразуем в 0/0 и применим несколько раз правило Бернулли-Лопиталя
![=e^{ lim_{n to infty} ( frac{1}{ frac{1}{2x+1} } ln ( frac{x+3}{x-2} ))}= \ \
=e^{ lim_{n to infty} ( frac{ ln ( frac{x+3}{x-2} )}{ frac{1}{2x+1} })}= \ \
=e^{ lim_{n to infty} frac{ (ln ( frac{x+3}{x-2} ))'}{( frac{1}{2x+1})' }}= (*)\ \
(ln ( frac{x+3}{x-2} ))'= frac{x-2}{x+3}* ( frac{x+3}{x-2} )'=
frac{x-2}{x+3}* frac{(x-2)-(x+3)}{(x-2)^{2}} =- frac{5}{(x-2)(x+3)} ; \ \
((2x+1)^{-1})'=-(2x+1)^{-2}*(2x+1)'= -frac{2}{(2x+1)^{2}} \ \](https://tex.z-dn.net/?f=%3De%5E%7B+lim_%7Bn+to+infty%7D+%28+frac%7B1%7D%7B+frac%7B1%7D%7B2x%2B1%7D+%7D+ln+%28+frac%7Bx%2B3%7D%7Bx-2%7D+%29%29%7D%3D+%5C++%5C+%0A%3De%5E%7B+lim_%7Bn+to+infty%7D+%28+frac%7B+ln+%28+frac%7Bx%2B3%7D%7Bx-2%7D+%29%7D%7B+frac%7B1%7D%7B2x%2B1%7D+%7D%29%7D%3D+%5C++%5C+%0A%3De%5E%7B+lim_%7Bn+to+infty%7D++frac%7B+%28ln+%28+frac%7Bx%2B3%7D%7Bx-2%7D+%29%29%27%7D%7B%28+frac%7B1%7D%7B2x%2B1%7D%29%27+%7D%7D%3D++%28%2A%29%5C++%5C+%0A%28ln+%28+frac%7Bx%2B3%7D%7Bx-2%7D+%29%29%27%3D+frac%7Bx-2%7D%7Bx%2B3%7D%2A++%28+frac%7Bx%2B3%7D%7Bx-2%7D+%29%27%3D%0Afrac%7Bx-2%7D%7Bx%2B3%7D%2A+frac%7B%28x-2%29-%28x%2B3%29%7D%7B%28x-2%29%5E%7B2%7D%7D+%3D-+frac%7B5%7D%7B%28x-2%29%28x%2B3%29%7D+%3B+%5C++%5C+%0A%28%282x%2B1%29%5E%7B-1%7D%29%27%3D-%282x%2B1%29%5E%7B-2%7D%2A%282x%2B1%29%27%3D+-frac%7B2%7D%7B%282x%2B1%29%5E%7B2%7D%7D++%5C++%5C+%0A "=e^{ lim_{n to infty} ( frac{1}{ frac{1}{2x+1} } ln ( frac{x+3}{x-2} ))}= \ \
=e^{ lim_{n to infty} ( frac{ ln ( frac{x+3}{x-2} )}{ frac{1}{2x+1} })}= \ \
=e^{ lim_{n to infty} frac{ (ln ( frac{x+3}{x-2} ))'}{( frac{1}{2x+1})' }}= (*)\ \
(ln ( frac{x+3}{x-2} ))'= frac{x-2}{x+3}* ( frac{x+3}{x-2} )'=
frac{x-2}{x+3}* frac{(x-2)-(x+3)}{(x-2)^{2}} =- frac{5}{(x-2)(x+3)} ; \ \
((2x+1)^{-1})'=-(2x+1)^{-2}*(2x+1)'= -frac{2}{(2x+1)^{2}} \ \")
![(*)=e^{ lim_{n to infty} frac{ frac{5}{(x-2)(x+3)} }{ frac{2}{(2x+1)^{2}} }}= \ \
=e^{ lim_{n to infty} frac{5(2x+1)^{2}}{2(x-2)(x+3)} }= \ \
=e^{ lim_{n to infty} frac{5(4 x^{2} +4x+1)}{2(x^{2}+x-6)} }= \ \
=e^{ lim_{n to infty} frac{(5(4 x^{2} +4x+1))'}{(2(x^{2}+x-6))'} }= \ \
=e^{ lim_{n to infty} frac{5(8 x +4)}{2(2x+1)} }= \ \
=e^{ lim_{n to infty} frac{5(8 x +4)'}{2(2x+1)'} }= \ \
=e^{ lim_{n to infty} frac{5*8}{2*2} }=](https://tex.z-dn.net/?f=%28%2A%29%3De%5E%7B+lim_%7Bn+to+infty%7D++frac%7B+frac%7B5%7D%7B%28x-2%29%28x%2B3%29%7D+%7D%7B+frac%7B2%7D%7B%282x%2B1%29%5E%7B2%7D%7D+%7D%7D%3D+%5C++%5C+%0A+%3De%5E%7B+lim_%7Bn+to+infty%7D+++frac%7B5%282x%2B1%29%5E%7B2%7D%7D%7B2%28x-2%29%28x%2B3%29%7D+%7D%3D+%5C++%5C+%0A+%3De%5E%7B+lim_%7Bn+to+infty%7D+++frac%7B5%284+x%5E%7B2%7D+%2B4x%2B1%29%7D%7B2%28x%5E%7B2%7D%2Bx-6%29%7D+%7D%3D+%5C++%5C+%0A+%3De%5E%7B+lim_%7Bn+to+infty%7D+++frac%7B%285%284+x%5E%7B2%7D+%2B4x%2B1%29%29%27%7D%7B%282%28x%5E%7B2%7D%2Bx-6%29%29%27%7D+%7D%3D+%5C++%5C%0A++%3De%5E%7B+lim_%7Bn+to+infty%7D+++frac%7B5%288+x+%2B4%29%7D%7B2%282x%2B1%29%7D+%7D%3D+%5C++%5C+%0A++%3De%5E%7B+lim_%7Bn+to+infty%7D+++frac%7B5%288+x+%2B4%29%27%7D%7B2%282x%2B1%29%27%7D+%7D%3D+%5C++%5C+%0A%3De%5E%7B+lim_%7Bn+to+infty%7D+++frac%7B5%2A8%7D%7B2%2A2%7D+%7D%3D "(*)=e^{ lim_{n to infty} frac{ frac{5}{(x-2)(x+3)} }{ frac{2}{(2x+1)^{2}} }}= \ \
=e^{ lim_{n to infty} frac{5(2x+1)^{2}}{2(x-2)(x+3)} }= \ \
=e^{ lim_{n to infty} frac{5(4 x^{2} +4x+1)}{2(x^{2}+x-6)} }= \ \
=e^{ lim_{n to infty} frac{(5(4 x^{2} +4x+1))'}{(2(x^{2}+x-6))'} }= \ \
=e^{ lim_{n to infty} frac{5(8 x +4)}{2(2x+1)} }= \ \
=e^{ lim_{n to infty} frac{5(8 x +4)'}{2(2x+1)'} }= \ \
=e^{ lim_{n to infty} frac{5*8}{2*2} }=")
![=e^{ lim_{n to infty} frac{40}{4}}= \ \
=e^ lim_{n to infty} 10=](https://tex.z-dn.net/?f=%3De%5E%7B+lim_%7Bn+to+infty%7D++frac%7B40%7D%7B4%7D%7D%3D+%5C++%5C+%0A%3De%5E+lim_%7Bn+to+infty%7D+10%3D "=e^{ lim_{n to infty} frac{40}{4}}= \ \
=e^ lim_{n to infty} 10=")
под пределом получили неопределенность 0*∞ , преобразуем в 0/0 и применим несколько раз правило Бернулли-Лопиталя
Новые вопросы