1003047409

divergent and $\underset{n\to \mathrm{\infty }}{lim}\frac{2\cdot 3^n+4^n+5}{4\cdot 3^n-1}=\mathrm{\infty }$

convergent and $\underset{n\to \mathrm{\infty }}{lim}\frac{2\cdot 3^n+4^n+5}{4\cdot 3^n-1}=\frac{1}{2}$

convergent and $\underset{n\to \mathrm{\infty }}{lim}\frac{2\cdot 3^n+4^n+5}{4\cdot 3^n-1}=\frac{1}{4}$

convergent and $\underset{n\to \mathrm{\infty }}{lim}\frac{2\cdot 3^n+4^n+5}{4\cdot 3^n-1}=0$

divergent and it does not have an infinite limit