Find Limit of $$\frac{x}{e} - \left(\frac{x}{x+1}\right)^x$$ at Infinity

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Discussion Overview

The discussion centers around finding the limit of the expression $$\lim_{x\to \infty} x\left[\frac{1}{e} - \left(\frac{x}{x+1}\right)^x\right]$$. The scope includes mathematical reasoning and limit evaluation as x approaches infinity.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant presents the limit to be evaluated, seeking insights on its behavior as x approaches infinity.
  • Another participant acknowledges a mistake regarding the sign in the expansion of the logarithm, indicating a potential error in their earlier reasoning.

Areas of Agreement / Disagreement

The discussion does not show clear agreement or resolution, as it primarily consists of a limit inquiry and an acknowledgment of a mistake without further elaboration on the limit itself.

Contextual Notes

There may be unresolved mathematical steps related to the limit evaluation and the implications of the logarithmic expansion error.

Euge
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Find the limit $$\lim_{x\to \infty} x\left[\frac{1}{e} - \left(\frac{x}{x+1}\right)^x\right]$$
 
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Substiuting x=1/y, the task is
\lim_{y\rightarrow +0}\frac{e^{-1}-(1+y)^{-1/y}}{y}
Considering
-\frac{1}{y}\ln(1+y) \approx -\frac{1}{y} (y+\frac{y^2}{2})
the task is
\lim_{y\rightarrow +0}\frac{e^{-1}(1-e^{-y/2})}{y}=(2e)^{-1}
[EDIT]
-\frac{1}{y}\ln(1+y) \approx -\frac{1}{y} (y-\frac{y^2}{2})
the task is
\lim_{y\rightarrow +0}\frac{e^{-1}(1-e^{y/2})}{y}=-(2e)^{-1}
 
Last edited:
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My bad, wrong sign in expansion of log.
 
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