求数列$a_n=\left(\frac{n-4}{n}\right)^{1-n}$的极限：推导存疑

Hey there! You’ve already nailed the initial rewrite of the sequence—nice work getting to $\frac{\left(1+\frac{-4}{n}\right)}{\left(1+\frac{-4}{n}\right)^n}$. Let’s walk through the rest of the process step by step, using key limit rules and a fundamental calculus identity.

Key Background Rule #

First, recall this critical limit that’s essential for problems like this:

$\lim_{n \to \infty} \left(1 + \frac{k}{n}\right)^n = e^k$
This holds for any constant $k$, and it’s the backbone of solving the denominator part of your expression.

Step 1: Split the Limit Using the Quotient Rule #

Since both the numerator and denominator have valid limits as $n \to \infty$ (and the denominator’s limit isn’t zero), we can apply the limit quotient rule:
$\lim_{n \to \infty} \frac{f(n)}{g(n)} = \frac{\lim_{n \to \infty} f(n)}{\lim_{n \to \infty} g(n)}$

Applying this to your expression:
$\lim_{n \to \infty} \frac{\left(1+\frac{-4}{n}\right)}{\left(1+\frac{-4}{n}\right)^n} = \frac{\lim_{n \to \infty} \left(1+\frac{-4}{n}\right)}{\lim_{n \to \infty} \left(1+\frac{-4}{n}\right)^n}$

Step 2: Evaluate the Numerator’s Limit #

As $n$ approaches infinity, the term $\frac{-4}{n}$ gets closer and closer to 0. So:
$\lim_{n \to \infty} \left(1+\frac{-4}{n}\right) = 1 + 0 = 1$

Step 3: Evaluate the Denominator’s Limit #

Using the key limit rule we mentioned earlier, set $k = -4$. Then:
$\lim_{n \to \infty} \left(1+\frac{-4}{n}\right)^n = e^{-4}$
Remember that $e^{-4}$ is the same as $\frac{1}{e^4}$.

Step 4: Combine the Results #

Now plug the numerator and denominator limits back into the quotient:
$\frac{1}{e^{-4}} = 1 \times e^4 = e^4$

So the final limit of the sequence $a_n$ is $e^4$.

内容的提问来源于stack exchange，提问作者Finn Eggers