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\lim_{n\rightarrow \infty}(1+\frac{1}{-n})^{-n}=e

Creado el: 13 de julio de 2025

Respondido usando O3 por Chat01

Pregunta

\lim_{n\rightarrow \infty}(1+\frac{1}{-n})^{-n}=e

Pensamiento

Clarifying limit expression

The user typed an expression involving a limit, lim⁡n→∞(1+1−n)−n=e\lim_{n\rightarrow \infty}(1+\frac{1}{-n})^{-n}=e. It looks like they're referring to the standard limit for ee, but with a negative sign in the denominator and exponent. After breaking it down, it seems both forms, (1−1n)−n(1 - \frac{1}{n})^{-n} and (1+1−n)−n(1 + \frac{1}{-n})^{-n}, give the same result. This limit tends to ee, so the expression indeed equals ee, but might be a bit ambiguous in how it’s framed.

Thought for 7 seconds

Respuesta

Yep — that limit really does land on ee.
Here’s the quick way to see it:

lim⁡n→∞(1+1−n)−n=lim⁡n→∞(1−1n)−n.\lim_{n\to\infty}\bigl(1+\tfrac1{-n}\bigr)^{-n} =\lim_{n\to\infty}\bigl(1-\tfrac1n\bigr)^{-n}.

Take natural logs to turn the power into a product:

ln⁡((1−1n)−n)=−n  ln⁡ ⁣(1−1n).\ln\Bigl(\bigl(1-\tfrac1n\bigr)^{-n}\Bigr) =-n\;\ln\!\bigl(1-\tfrac1n\bigr).

Now expand the logarithm for small xx (ln⁡(1−x)=−x−x22−…\ln(1-x)= -x-\frac{x^{2}}{2}-\dots) with x=1nx=\tfrac1n:

−n(−1n−12n2+O(1n3))=1+12n+O(1n2).-n\bigl(-\tfrac1n-\tfrac1{2n^{2}}+O(\tfrac1{n^{3}})\bigr) =1+\tfrac1{2n}+O(\tfrac1{n^{2}}).

As n→∞n\to\infty, that expression tends to 11. Exponentiating again gives

exp⁡ ⁣(1+12n+O(1n2))  ⟶  e.\exp\!\bigl(1+\tfrac1{2n}+O(\tfrac1{n^{2}})\bigr)\;\longrightarrow\;e.

So

lim⁡n→∞(1−1n)−n=e.\boxed{\displaystyle\lim_{n\to\infty}\bigl(1-\tfrac1n\bigr)^{-n}=e.}

(If you instead keep the usual positive sign in both places, (1+1n)n→e\bigl(1+\tfrac1n\bigr)^{n}\to e as you may already know. Flipping both the sign inside and the sign on the exponent lands you at the same constant.)

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