Vali
- 48
- 0
Why $$\lim_{n\rightarrow \infty }\frac{n^{2016}\cdot 2^{n-1}}{3^{n}}=0$$ ?
Because $3^{n}> 2^{n-1} $ ?
Because $3^{n}> 2^{n-1} $ ?
The discussion centers on evaluating the limit $$\lim_{n\rightarrow \infty }\frac{n^{2016}\cdot 2^{n-1}}{3^{n}}$$ and understanding the behavior of the terms involved as \( n \) approaches infinity. Participants explore the dominance of exponential functions over polynomial functions in this context.
Participants generally agree on the idea that exponential functions dominate polynomial functions in limits as \( n \) approaches infinity. However, there is some confusion regarding the specific forms of the expressions and how to apply the concepts correctly.
Participants express uncertainty about the application of the limit properties and the identification of terms in the context of exponential and polynomial growth. There are unresolved questions regarding the specific forms of the expressions used in the limit evaluation.
Vali said:I understood the simplification but I don't understand the form.I mean, I know that $\frac{n^b}{a^n}$ tends to 0 but in my form I have $n^{b}*a^{n}$.
In $\frac{n^b}{a^n}$ which is $n^{b}$ and which is $a^{n}$ ?