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Please briefly explain why you feel this question should be reported.
Va rugam explicate, pe scurt, de ce credeti ca aceasta intrebare trebuie raportata.
Motivul pentru care raportezi utilizatorul.
Am urmatorul sir:
![\[ a_n = \left( {\frac{n}{{\sqrt {n^2 + 2} }}} \right)^{n^2 } \]](https://anidescoala.ro/wp-content/ql-cache/quicklatex.com-7ec4122d718bd38c91c5263f60ae4dea_l3.png "formula matematica")
Imi cere sa ii calculez limita.
Voi aplica teorema conform careia:
![\[ {\lim }\limits_{n \to \infty } \left( {1 + x_n } \right)^{\frac{1}{{x_n }}} = e\,\,\,\,\,\,unde\,\,x_n \to 0 \]](https://anidescoala.ro/wp-content/ql-cache/quicklatex.com-f347005060c2c1c6f02ca59c172d9f89_l3.png "formula matematica")
Avem asa:
![\[ {\lim }\limits_{n \to \infty } \left( {\frac{n}{{\sqrt {n^2 + 2} }}} \right)^{n^2 } = {\lim }\limits_{n \to \infty } \left( {1 + \frac{{n - \sqrt {n^2 + 2} }}{{\sqrt {n^2 + 2} }}} \right)^{n^2 } \]](https://anidescoala.ro/wp-content/ql-cache/quicklatex.com-b2b7b8c079ea59a9b14b21841f0a12d6_l3.png "formula matematica")
Acum o sa demonstrez ca limita sirului
![\[ {\frac{{n - \sqrt {n^2 + 2} }}{{\sqrt {n^2 + 2} }}} \]](https://anidescoala.ro/wp-content/ql-cache/quicklatex.com-ca0b5a539debfd3aa22ed5ed97200f86_l3.png "formula matematica")
este de fapt zero:
![\[ \begin{array}{l} {\lim }\limits_{n \to \infty } \frac{{n - \sqrt {n^2 + 2} }}{{\sqrt {n^2 + 2} }} = {\lim }\limits_{n \to \infty } \frac{{n - n\sqrt {1 + \frac{2}{{n^2 }}} }}{{n\sqrt {1 + \frac{2}{{n^2 }}} }} = {\lim }\limits_{n \to \infty } \frac{{n\left( {1 - \sqrt {1 + \frac{2}{{n^2 }}} } \right)}}{{n\sqrt {1 + \frac{2}{{n^2 }}} }} = \\ = {\lim }\limits_{n \to \infty } \frac{{1 - \sqrt {1 + \frac{2}{{n^2 }}} }}{{\sqrt {1 + \frac{2}{{n^2 }}} }} = \frac{{1 - 1}}{1} = 0 \\ \end{array} \]](https://anidescoala.ro/wp-content/ql-cache/quicklatex.com-2afa41f66b62791d8a0fb8ea83c1eb4a_l3.png "formula matematica")
Acestea fiind scrise, revenim la exercitiu:
![\[ \begin{array}{l} {\lim }\limits_{n \to \infty } \left( {1 + \frac{{n - \sqrt {n^2 + 2} }}{{\sqrt {n^2 + 2} }}} \right)^{n^2 } = {\lim }\limits_{n \to \infty } \left[ {\left\{ {\left[ {\left( {1 + \frac{{n - \sqrt {n^2 + 2} }}{{\sqrt {n^2 + 2} }}} \right)^{\frac{{\sqrt {n^2 + 2} }}{{n - \sqrt {n^2 + 2} }}} } \right]^{\frac{{n - \sqrt {n^2 + 2} }}{{\sqrt {n^2 + 2} }}} } \right\}} \right]^{n^2 } = e^{ {\lim }\limits_{n \to \infty } \frac{{n^2 \left( {n - \sqrt {n^2 + 2} } \right)}}{{\sqrt {n^2 + 2} }}} = \\ = e^{ {\lim }\limits_{n \to \infty } \frac{{n^2 \left( {n - n\sqrt {1 + \frac{2}{{n^2 }}} } \right)}}{{n\sqrt {1 + \frac{2}{{n^2 }}} }}} = e^{ {\lim }\limits_{n \to \infty } \frac{{n^2 \left( {1 - \sqrt {1 + \frac{2}{{n^2 }}} } \right)}}{{\sqrt {1 + \frac{2}{{n^2 }}} }}} = e^{\frac{{\infty \cdot \left( {1 - 1} \right)}}{1}} \\ \end{array} \]](https://anidescoala.ro/wp-content/ql-cache/quicklatex.com-a85f7e4c4bc4270cb31f52abe3a7d4f5_l3.png "formula matematica")
Ajung la nedeterminarea 
. Cum pot scapa de asta?
Multumesc.
Vei amplifica fractia cu conjugata parantezei ,adica cu n+rad(n^2+2)
Multumesc cu respect sandy_sc. Se pare ca merge.