Limit Function: Finding Solution

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    Function Limit
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Discussion Overview

The discussion revolves around finding the limit of the function $$\lim_{{n}\to{\infty}} \frac{n^n}{({n + 1})^{n + 1}}$$. Participants explore different methods and approaches to simplify and evaluate this limit, including the use of L'Hospital's Rule and exponential limits.

Discussion Character

  • Mathematical reasoning, Technical explanation, Debate/contested

Main Points Raised

  • Some participants simplify the limit to $$\lim_{{n}\to{\infty}} \frac{n^n}{({n + 1})^{n}(n + 1)}$$ but express uncertainty about how to proceed further.
  • One participant suggests using L'Hospital's Rule to evaluate the limit, providing a detailed breakdown of the steps involved.
  • Another participant proposes an alternative approach using the limit property $$\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n=e^x$$ to arrive at a limit of $$e^{-1}$$.
  • A different perspective is presented, arguing that the limit approaches 0 by bounding the expression and showing that $$\frac{n^n}{({n + 1})^{n + 1}} \le 1^n \cdot \frac{1}{n + 1}$$ as $$n$$ approaches infinity.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the limit's value, as multiple competing views are presented, including approaches that suggest the limit is 0 and others that suggest it approaches $$e^{-1}$$.

Contextual Notes

Some methods rely on specific mathematical properties and assumptions that may not be universally accepted or applicable in all contexts, leaving certain steps unresolved.

tmt1
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I'm trying to find the limit of this function:

$$\lim_{{n}\to{\infty}} \frac{n^n}{({n + 1})^{n + 1}}$$

I can simplify it to this:

$$\lim_{{n}\to{\infty}} \frac{n^n}{({n + 1})^{n}(n + 1)}$$

But I'm not sure of the best way to proceed.
 
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tmt said:
I'm trying to find the limit of this function:

$$\lim_{{n}\to{\infty}} \frac{n^n}{({n + 1})^{n + 1}}$$

I can simplify it to this:

$$\lim_{{n}\to{\infty}} \frac{n^n}{({n + 1})^{n}(n + 1)}$$

But I'm not sure of the best way to proceed.

$\displaystyle \begin{align*} &= \lim_{n \to \infty} \left( \frac{n}{n + 1} \right) ^n \left( \frac{1}{n + 1} \right) \\ &= \lim_{n \to \infty} \left( \frac{n}{n + 1} \right) ^n \, \lim_{n \to \infty} \frac{1}{n + 1} \\ &= \lim_{n \to \infty} \left( 1 - \frac{1}{n + 1} \right) ^n \,\lim_{n \to \infty} \frac{1}{n + 1} \\ &= \lim_{n \to \infty} \mathrm{e}^{ \ln{ \left[ \left( 1 - \frac{1}{n + 1} \right) ^n \right] } }\,\lim_{n \to \infty} \frac{1}{n + 1} \\ &= \mathrm{e}^{ \lim_{n \to \infty} \frac{\ln{ \left( 1 - \frac{1}{n + 1} \right) }}{\frac{1}{n}} }\,\lim_{n \to \infty} \frac{1}{n + 1} \end{align*}$

Use L'Hospital's Rule to finish it off :)
 
An alternative to L'Hospital's rule in this context is the fact that $\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n=e^x$. Then
\begin{align}
\lim_{n\to\infty}\left(1-\frac{1}{n+1}\right)^n&=
\lim_{n\to\infty}\left(1-\frac{1}{n+1}\right)^{n+1}\left(\frac{n+1}{n}\right)\\
&=\lim_{n\to\infty}\left(1-\frac{1}{n}\right)^{n}\lim_{n\to\infty}\left(\frac{n+1}{n}\right)\\
&=e^{-1}\cdot1.
\end{align}
 
Alternatively:
$$\frac{n^n}{({n + 1})^{n + 1}} = \left(\frac{n}{n + 1}\right)^{n} \cdot \frac{1}{n + 1} \le 1^n \cdot \frac{1}{n + 1} \to 0
$$
So:
$$\lim_{{n}\to{\infty}} \frac{n^n}{({n + 1})^{n + 1}} = 0$$
 

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