In summary, The limit of x/e - (x/(x+1))^x at infinity is 1/e. To find the limit of a function at infinity, you can use the rules of limits such as factoring, simplifying, and applying L'Hopital's rule if necessary. The limit of this function at infinity is equal to 1/e because as x approaches infinity, the terms x and x+1 become negligible compared to x/(x+1), which approaches 1. Additionally, the term (1+1/x)^x approaches e as x approaches infinity. It is not possible for the limit of this function at infinity to be a different value, as both the function and its derivative approach the same value as x
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Find the limit $$\lim_{x\to \infty} x\left[\frac{1}{e} - \left(\frac{x}{x+1}\right)^x\right]$$
Substiuting x=1/y, the task is
[tex]\lim_{y\rightarrow +0}\frac{e^{-1}-(1+y)^{-1/y}}{y}[/tex]
Considering
[tex]-\frac{1}{y}\ln(1+y) \approx -\frac{1}{y} (y+\frac{y^2}{2})[/tex]
the task is
[tex]\lim_{y\rightarrow +0}\frac{e^{-1}(1-e^{-y/2})}{y}=(2e)^{-1}[/tex]
[EDIT]
[tex]-\frac{1}{y}\ln(1+y) \approx -\frac{1}{y} (y-\frac{y^2}{2})[/tex]
the task is
[tex]\lim_{y\rightarrow +0}\frac{e^{-1}(1-e^{y/2})}{y}=-(2e)^{-1}[/tex]