Question

Calculati limita sirului $x_{n}= \frac{2^n+a^n}{3^n+4^n}$, a>0.

Am nevoie de confirmare ca lim=0 pentru a<4 si lim=infinit pentru a>4 si calularea limitei sirului $x_{n}= \frac{2^n+4^n}{3^n+4^n}$. Pentru ca mie imi da lim=1 si sirul e descrescator cu primul termen 6/7. Mersi

Answer 1

$\lim\limits_{n \to \infty} \frac{2^n+a^n}{3^n+4^n}=\lim\limits_{n\to\infty} (\frac{2^n}{3^n+4^n}+ \frac{a^n}{3^n+4^n})= \\ \lim\limits_{n\to\infty} \frac{2^n}{3^n+4^n}+ \lim\limits_{n\to\infty}\frac{a^n}{3^n+4^n}=0+\lim\limits_{n\to\infty}\frac{a^n}{3^n+4^n} \\ =\lim\limits_{n\to\infty}\frac{a^n}{3^n+4^n}$
$\lim\limits_{n\to\infty\atop a\ \textgreater \ 4} \frac{a^n}{3^n+4^n} =\lim\limits_{n\to\infty\atop a\ \textgreater \ 4} \frac{a^n}{a^n( \frac{3^n}{a^n} + \frac{4^n}{a^n}) } =\lim\limits_{n\to\infty\atop a\ \textgreater \ 4} \frac{1}{0_++0_+} =\infty \\ Deci\ l=0\ pt\ a\ \textless \ 4\ si\ \infty\ pt\ a\ \textgreater \ 4.$