S:L15.231

S:L15.231 (Andrei Horvat-Marc)

Fie Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left(a_n\right)_{n\ge 1}} un șir crescător de numere reale strict pozitive cu Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim\limits_{n\to \infty} a_n = a} . Arătați că Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim\limits_{n\to\infty} \dfrac{a-a_n}{\ln\frac{a}{a_n}} = a} Soluție

Fie $n\in \mathbb {N} ^{\ast }$. Se consideră funcția $f_{n}:\left[a_{n},a\right]\to \mathbb {R}$, cu $f_{n}\left(x\right)=\ln x$. Pentru această funcție se aplică teorema lui Lagrange, deci există $c_{n}\in \left(a_{n},a\right)$ astfel încât

$${\frac {\ln a-\ln a_{n}}{a-a_{n}}}={\frac {1}{c_{n}}}.$$

Cum $\lim \limits _{n\to \infty }c_{n}=a$ se obține Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim\limits_{n\to\infty} \dfrac{a-a_n}{\ln\frac{a}{a_n}} = \lim\limits_{n\to \infty} c_n = a.}