Why Does $$\lim_{n\rightarrow \infty }\frac{n^{2016}\cdot 2^{n-1}}{3^{n}}=0$$?

• MHB
• Vali
In summary, the limit of $\frac{n^b}{a^n}$ as n goes to infinity is 0, and the limit of $n^{2016}\cdot 2^{n-1}}{3^{n}}$ is $n^{2016}\cdot \frac{2^n}{3^n}.$
Vali
Why $$\lim_{n\rightarrow \infty }\frac{n^{2016}\cdot 2^{n-1}}{3^{n}}=0$$ ?
Because $3^{n}> 2^{n-1}$ ?

Not that alone. $$\frac{2^{n-1}}{3^n}= \frac{1}{2}\left(\frac{2}{3}\right)^n$$ as well as the fact that $$a^n$$ "dominates" $$n^b$$ as n goes to infinity. That is, for any a and b, larger than 1, the limit of $$\frac{n^b}{a^n}$$, as n goes to infinity, is 0.

Thank you for the response!
Yes, I know that log<power<exponential<factorial but it's not completely clear for me in this case.I have exponential function at denominator and numerator too.That $2^{n-1}$ is exponential but $3^{n}$ is also an exponential function but is higher than the first one.I'm a little bit confused..

That's why we simplify it as:
$$\lim_{n\rightarrow \infty }\frac{n^{2016}\cdot 2^{n-1}}{3^{n}} =\lim_{n\rightarrow \infty }n^{2016}\cdot \frac{ 2^{n-1}}{3^{n}} =\lim_{n\rightarrow \infty }n^{2016}\cdot \frac 2 2 \cdot \frac{ 2^{n-1}}{3^{n}} =\lim_{n\rightarrow \infty }n^{2016}\cdot \frac 1 2 \cdot \frac{ 2^{n}}{3^{n}} =\lim_{n\rightarrow \infty }\frac 1 2 \cdot n^{2016}\cdot \left(\frac{ 2}{3}\right)^n$$
Now we can use that domination order as Country Boy explained, can't we?

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 $a^{n}$ ?

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 $a^{n}$ ?

We can rewrite what we have as:
$$\lim_{n\rightarrow \infty }\frac 1 2 \cdot n^{2016}\cdot \frac{2^n}{3^n} = \lim_{n\rightarrow \infty }\frac 1 2 \cdot n^{2016}\cdot \frac{1}{\frac{3^n}{2^n}} = \lim_{n\rightarrow \infty }\frac 1 2 \cdot \frac{n^{2016}}{\left(\frac{3}{2}\right)^n}$$
Can we tell now which is $n^{b}$ and which is $a^{n}$ ?

I finally understood!
Thanks a lot!

1. What does the expression $\lim_{n\rightarrow \infty }\frac{n^{2016}\cdot 2^{n-1}}{3^{n}}$ represent?

The expression represents the limit of a sequence as n approaches infinity. It is the value that the sequence approaches as n gets larger and larger.

2. Why does the expression approach 0 as n gets larger?

This is because as n gets larger, the exponential term with the largest base (3^n) grows much faster than the other terms. This causes the overall fraction to approach 0.

3. How does the exponent of 2016 affect the limit?

The exponent of 2016 does not significantly affect the limit. It only adds a constant factor to the overall fraction, but as n gets larger, this becomes negligible in comparison to the exponential term.

4. Is this limit always equal to 0?

Yes, this limit will always approach 0 as n gets larger. This is because the exponential term with the largest base will always grow faster, making the overall fraction approach 0.

5. Can this limit be evaluated using L'Hopital's rule?

No, L'Hopital's rule can only be applied to limits of the form $\frac{f(x)}{g(x)}$ where both f(x) and g(x) approach 0 or infinity. In this case, the limit is not in that form, so L'Hopital's rule cannot be used.

Replies
2
Views
1K
Replies
2
Views
1K
Replies
3
Views
1K
Replies
2
Views
2K
Replies
9
Views
2K
Replies
16
Views
3K
Replies
2
Views
1K
Replies
17
Views
3K
Replies
2
Views
1K
Replies
11
Views
2K