- #1

Vali

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Because $3^{n}> 2^{n-1} $ ?

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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 $n^{2016}\cdot \frac{2^n}{3^n}.$

- #1

Vali

- 48

- 0

Because $3^{n}> 2^{n-1} $ ?

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- #2

HOI

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- #3

Vali

- 48

- 0

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..

- #4

I like Serena

Homework Helper

MHB

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$$\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?

- #5

Vali

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- 0

In $\frac{n^b}{a^n}$ which is $n^{b}$ and which is $a^{n}$ ?

- #6

I like Serena

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MHB

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Vali said:

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 $a^{n}$ ?

- #7

Vali

- 48

- 0

I finally understood!

Thanks a lot!

Thanks a lot!

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.

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.

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.

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.

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.

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