What is the limit of (n/(n-1))^(n+2) as n approaches infinity?

  • Thread starter Telemachus
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In summary, the conversation discusses a difficulty with solving a limit and the use of L'Hopital's rule to simplify it. The speaker also mentions an alternate approach using the function f(x)=(x/(x-1))^(x+2) and applying limit laws to solve the limit.
  • #1
Telemachus
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Homework Statement


Hi there. I've found some difficulties on solving this limit:
[tex]\displaystyle\lim_{n \to{}\infty}{(\displaystyle\frac{n}{n-1})^{n+2}}[/tex]

I thought of working with the function [tex]f(x)=(\displaystyle\frac{x}{x-1})^{x+2}[/tex]
And then apply L'Hopital

This way:
[tex]\displaystyle\lim_{x \to{}\infty}{(\displaystyle\frac{x}{x-1})^{x+2}=e^{\displaystyle\lim_{x \to{}\infty}{(x+2) ln (\displaystyle\frac{x}{x-1})}}=e^{\displaystyle\lim_{x \to{}\infty}{\displaystyle\frac{(\displaystyle\frac{x}{x-1})}{\displaystyle\frac{1}{(x+2)}}}[/tex]

And then I've applied L'hopital, but it didn't make the things easier. I've applied L'hopital unless two times. I don't know if what I did is right. And I think there must be an easier way of solving this.
 
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  • #2
Note that we have ...

[tex]\frac{n}{n-1} = 1 + \frac{1}{n-1}[/tex]

This gives us ...

[tex]\left(\frac{n}{n-1}\right)^{n+2} = \left(1+\frac{1}{n-1}\right)^{n-1}\left(1+\frac{1}{n-1}\right)^3[/tex]

Evaluate the limit as [itex]n \to \infty[/itex] for each of the terms on right and then apply some "limit laws" to justify your result.
 
  • #3
Thanks.
 

1. What is the concept of "Indetermination 1^infinity"?

"Indetermination 1^infinity" refers to the mathematical concept of an indeterminate form, where the value of a limit cannot be determined solely by examining the expression. It involves raising the number 1 to the power of infinity, which results in an infinitely large number.

2. What are some examples of indeterminate forms involving 1^infinity?

Some examples of indeterminate forms involving 1^infinity include 1^∞, 0^∞, and ∞^0. These expressions cannot be evaluated without further information or context, as the result can vary depending on the specific values involved.

3. How is "Indetermination 1^infinity" used in mathematics?

"Indetermination 1^infinity" is often encountered in calculus, specifically in limits involving exponential and logarithmic functions. It is used to determine the behavior of a function as it approaches a specific point or value.

4. Can "Indetermination 1^infinity" ever have a defined value?

No, "Indetermination 1^infinity" is considered to be undefined or indeterminate. This means that the value cannot be determined without additional information or context.

5. How do mathematicians handle indeterminate forms involving 1^infinity?

To evaluate indeterminate forms involving 1^infinity, mathematicians use techniques such as L'Hopital's rule, which involves taking the derivatives of the numerator and denominator of the expression. They may also use algebraic manipulation or substitution to simplify the expression before evaluating the limit.

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