Question

$\lim_{n \to \infty} \frac{ 2^{n}+ 3^{n} }{ 2^{n+1}+ 3^{n+1} }$     rezovati explicit!

Answer 1

cazul : ∞ / ∞ 
factor fortat  3 ( la  n ) , la numarator , la numitor 
stim ca subunitarul  ( 2 /3)  ( la n ) --------> 0                daca n-------->∞
ex = 1 /3¹ = 1 / 3 

Answer 2

$\displaystyle \lim_{n \to \infty} \frac{2^n+3^n}{2^{n+1}+3^{n+1}} = \lim_{n \to \infty} \frac{3^n \left( \frac{2}{3} \right)^n+1}{3^{n+1} \left( \frac{1}{2^{-n}\cdot 3^{n+1}} +1\right)} = \lim_{n \to \infty} \frac{\left( \frac{2}{3} \right)^n+1}{\left(3 \left( \frac{3}{2} \right)^{-n-1}+1\right)}$

$\displaystyle = \frac{ \lim_{n \to \infty}\left( \frac{2}{3} \right)^n+1}{ \lim_{n \to \infty} \left(3 \left( \frac{3}{2}\right)^{-n-1}+1 \right) }=\frac{ \lim_{n \to \infty}\left(\left( \frac{2}{3} \right)^n\right)+1}{ 3\left( \frac{1}{\infty } }\left( \frac{3}{2} \right)^{-1}+1 \right)}= \frac{0+1}{3} = \boxed{\frac{1}{3} }$