What Is the Limit of the Sequence as \( n \) Approaches Infinity?

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SUMMARY

The limit of the sequence as \( n \) approaches infinity for \( \lim_{n\rightarrow \infty} \frac{e^n + 3^n}{5^n} \) is determined by the growth rates of the exponential functions involved. The denominator \( 5^n \) grows faster than both \( e^n \) and \( 3^n \), leading to the conclusion that the limit is 0. L'Hôpital's rule is not necessary for this evaluation, as the comparison of growth rates suffices to demonstrate that the numerator's growth is outpaced by the denominator's growth. The Squeeze Theorem can also be applied to reinforce this conclusion.

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Homework Statement

Find the limit of the sequence

[tex]\lim n\rightarrow \infty \frac{e^n+3^n}{5^n}[/tex]

Homework Equations


The Attempt at a Solution



L'Hopital's rule never ends with this one. But even after taking the first derivative of the top and bottom it shows that 5x will always be grower faster than the combined derivatives of the top.

[tex]\frac{e^n + ln(3)3^n)}{ln(5)5^n}[/tex]

Assuming that I only pick values greater than 1, because this problem is a sequence, then I will always have the bottom value of the derivative greater than the combined values of the top. Does prove that the denominator is growing faster than the top? It's seems kinda hokey to just say that.
 
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No need L'Hopital Rule ... (Dont abandon ur basic tools when u are equipped with advanced tools)

[tex]\frac{e^{n}}{5^{n}} = (\frac{e}{5})^{n}[/tex]

What can u say abt the fraction as [tex]n\rightarrow\infty[/tex] ?

Likewise repeat the similar argument for 3/5
 
You have the right idea! and surely 5n grows faster than the other bases.
A nice way to show this is by the "squeeze theorem". i.e. bound the expression above and below by expressions that are a bit smaller and a bit larger and show they all go to the same limit, thus the "squeezed" expression has the same limit.
 

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