Exploring the Limits of $\lim_{n\rightarrow\infty}(\frac{n+1}{n})^{n+1}$

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In summary, both the limits of the product of 2 factors and the limit of a sequence with a negative x go to e.
  • #1
Nerpilis
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find the [tex]\lim_{n\rightarrow\infty}(\frac{n+1}{n})^{n+1} = [/tex]
so far what i have is
[tex]\lim_{n\rightarrow\infty}(\frac{n+1}{n})^{n+1} = \lim_{n\rightarrow\infty}(\frac{n}{n} + \frac{1}{n})^{n+1} = \lim_{n\rightarrow\infty}(1 + \frac{1}{n})^{n+1}\\[/tex]

I know this has got to go to [tex]e[/tex] or something very close, my question lies in if I should be concerned with the +1 in the n+1 in the exponet.
There is another one which I'm totaly stumped on for a starting point, my gut tells me the limit is 1 or it may not even exist altogether.
[tex]\lim_{n\rightarrow\infty}(\frac{n+3}{n+1})^{n+4} = [/tex]
but then again it could be +[tex]\infty[/tex] if you realize that in the () the numerator is larger than the denominator such that as the exponet is applied it goes to infinity instead od zero.
 
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  • #2
You said you already have a suspicion that the limit should have to do with e. So I assume you know the limit definition of e which has a very similar form?

The second has a very similar form. Can you rewrite it so the form resembles a limit you already know? (Like that of exp(x) or something).
 
  • #3
Did you notice that
[tex]\left(1+\frac{1}{n}\right)^{n+1}[/tex]
is the same as
[tex]\left(1+\frac{1}{n}\right)^n \left(1+ \frac{1}{n}\right)[/tex]
?
 
  • #4
no I did not realize that [tex]\left(1+\frac{1}{n}\right)^n \left(1+ \frac{1}{n}\right)[/tex] I guess that would make the limit equal to e^2?

as far as the second problem I don't know what exp(x) is but i recognized that I can get things to look more like e

[tex]\lim_{n\rightarrow\infty}(\frac{n+3}{n+1})^{n+4} = [/tex]

[tex]\lim_{n\rightarrow\infty}((\frac{n + 3}{n + 1})(\frac{1/n}{1/n}))^{n+4} = [/tex]
[tex]\lim_{n\rightarrow\infty}\frac{(1 + (3/n))^{n+4}}{(1 + (1/n))^{n+4}} =[/tex]


from here the denominator looks similar to e and the numerator almost looks like e, however I’m a little hesitant to call the limit e/(e^4) = 1/(e^3) therefore I applied bernouli’s inequality to the numerator :

[tex] \geq \lim_{n\rightarrow\infty}\frac{1 + (n + 4)(3/n)}{e^{4}} \geq \frac{\lim_{n\rightarrow\infty}(1 + 3 + (12/n))}{e^{4}} \geq \frac{4}{e^{4}}[/tex]

Is my thinking correct in operating this way?
 
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  • #5
Nerpilis said:
no I did not realize that [tex]\left(1+\frac{1}{n}\right)^n \left(1+ \frac{1}{n}\right)[/tex] I guess that would make the limit equal to e^2?
No. Why don't you just write it out and use some limit rules. This looks like the limit of a product of 2 factors, both with limits you know.

as far as the second problem I don't know what exp(x) is but i recognized that I can get things to look more like e
[tex]\lim_{n\rightarrow\infty}(\frac{n+3}{n+1})^{n+4} = [/tex]
[tex]\lim_{n\rightarrow\infty}((\frac{n + 3}{n + 1})(\frac{1/n}{1/n}))^{n+4} = [/tex]
[tex]\lim_{n\rightarrow\infty}\frac{(1 + (3/n))^{n+4}}{(1 + (1/n))^{n+4}} =[/tex]
from here the denominator looks similar to e and the numerator almost looks like e, however I’m a little hesitant to call the limit e/(e^4) = 1/(e^3)
Don't just guess what you think the limit should be. What makes you think that:
[tex]\lim_{n\to \infty} \left(1+\frac{1}{n}\right)^{n+4}=e^4[/tex]
? You can solve this limit in the same way as the previous one by considering it a product of 2 limits.

For a limit like: [itex]\lim_{n\to \infty} (1+x/n)^n[/itex] which looks a lot like the one for e, what method would be useful? (Hint: substitution)

Another way is to write:
[tex]\frac{n+3}{n+1}=\frac{n+1+2}{n+1}=1+\frac{2}{n+1}[/tex]
 
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  • #6
thank you , i have found my errors
[tex]\lim_{n\rightarrow\infty}(\frac{n+1}{n})^{n+1} = \lim_{n\rightarrow\infty}(\frac{n}{n} + \frac{1}{n})^{n+1} = \lim_{n\rightarrow\infty}(1 + \frac{1}{n})^{n+1}\\[/tex]
[tex]\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^n \left(1+ \frac{1}{n}\right) = e \times \lim_{n\rightarrow\infty}\left(1+ \frac{1}{n}\right) = e[/tex]
for the other one:
[tex]\lim_{n\rightarrow\infty}\left(\frac{n+3}{n+1}\right)^{n+4}= \lim_{n\to \infty} \left(1+\frac{2}{n+1}\right)^{n+4} = \lim_{n\rightarrow\infty}\left(1 + \frac{2}{n+1}\right)^{n+1}\left(1 + \frac{2}{n+1}\right)^{3} = \lim_{n\rightarrow\infty}\left(1 + \frac{2}{n+1}\right)^{n+1} \times 1[/tex]

now i have found that this actually goes to [tex]e^{2}[/tex] but i did get some help but i don't know how to prove it other than writing out the sequence values for e and then for this sequence [tex]\lim_{n\rightarrow\infty}\left(1 + \frac{2}{n+1}\right)^{n+1} [/tex]
and noticing that e is a subsequence of the above
 
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  • #7
You can make a substitution: m=n/x. That makes:

[tex]\lim_{n\to \infty} \left(1+\frac{x}{n}\right)^n[/tex]
look like:
[tex]\lim_{m\to \infty}\left(1+\frac{1}{m}\right)^{xm}[/tex]
Or m-> -infinity if x is negative.
 

1. What is the formula for calculating the limit of $\lim_{n\rightarrow\infty}(\frac{n+1}{n})^{n+1}$?

The formula for calculating the limit of $\lim_{n\rightarrow\infty}(\frac{n+1}{n})^{n+1}$ is $\lim_{n\rightarrow\infty}(\frac{n+1}{n})^{n+1} = e$ where $e$ is the mathematical constant approximately equal to 2.71828.

2. What does the limit of $\lim_{n\rightarrow\infty}(\frac{n+1}{n})^{n+1}$ represent?

The limit of $\lim_{n\rightarrow\infty}(\frac{n+1}{n})^{n+1}$ represents the maximum value that the expression $(\frac{n+1}{n})^{n+1}$ can approach as n approaches infinity.

3. How do you interpret the value of the limit of $\lim_{n\rightarrow\infty}(\frac{n+1}{n})^{n+1}$?

The value of the limit of $\lim_{n\rightarrow\infty}(\frac{n+1}{n})^{n+1}$ can be interpreted as the rate of growth or decay of a function as the input approaches infinity. In this case, the function is growing at a relatively slow rate, approaching the constant value of $e$.

4. What is the significance of exploring the limits of $\lim_{n\rightarrow\infty}(\frac{n+1}{n})^{n+1}$?

Exploring the limits of $\lim_{n\rightarrow\infty}(\frac{n+1}{n})^{n+1}$ allows us to better understand the behavior of functions and their rates of growth or decay as the input approaches infinity. It also has applications in fields such as calculus, where the concept of limits is crucial in solving complex mathematical problems.

5. How can the limit of $\lim_{n\rightarrow\infty}(\frac{n+1}{n})^{n+1}$ be calculated or approximated?

The limit of $\lim_{n\rightarrow\infty}(\frac{n+1}{n})^{n+1}$ can be calculated or approximated using various methods, such as numerical methods like the bisection method or the secant method, or algebraic methods like L'Hopital's rule. It can also be approximated by using a large value of n, such as n = 1000, to get a close estimate of the limit.

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