[tex]\lim\limits_{n \to \infty}\dfrac{2^n+3^n+a^n}{3^n+4^n}=\lim\limits_{n \to \infty}\dfrac{4^n\cdot \Big(\dfrac{2^n}{4^n}+\dfrac{3^n}{4^n}+\dfrac{a^n}{4^n}\Big)}{4^n\cdot \Big(\dfrac{3^n}{4^n}+1\Big)} =[/tex]
[tex]= \lim\limits_{n \to \infty}\dfrac{4^n\cdot \Bigg(\Big(\dfrac{2}{4}\Big)^n+\Big(\dfrac{3}{4}\Big)^n+\Big(\dfrac{a}{4}\Big)^n\Bigg)}{4^n\cdot \Bigg(\Big(\dfrac{3}{4}\Big)^n+1\Bigg)} = \\ \\\text{Toate fractiile mai mici decat 1 tind la 0, iar }4^n\text{ se simplifica.}\\ \\=\lim\limits_{n\to\infty}\Big(\dfrac{a}{4}\Big)^n = 0 \\ \\ \Rightarrow a\in(0,4),\quad \text{fiindca fractia trebuie sa fie }< 1[/tex]
