While calculating the following limit:
$$ \lim_{n \to \infty} \left(\frac{n-3}{n}\right)^n $$
I have used the following procedure:
\begin{align}
\lim_{n \to \infty} \left(\frac{n-3}{n}\right)^n =
\lim_{n \to \infty} \left(\frac{\frac{1}{n}\cdot(n-3)}{\frac{1}{n}\cdot n}\right)^n =
\lim_{n \to \infty} \left(1-\frac{3}{n}\right)^n = \\=
\lim_{n \to \infty} \left(1-\frac{3}{\infty}\right)^\infty =
\lim_{n \to \infty} \left(1-\frac{3}{\infty}\right)^\infty =
\lim_{n \to \infty} (1)^\infty = 1\\
\end{align}
I'm aware that the solution is $ e^{-3} $ but I'd like to know that rules I'm breaking in my process so the answer is wrong. I'm suspicious of the last two steps. I think that assuming that $ 3/\infty $ tends to 0 as $ n $ approaches $ \infty $ is fine and the result would approach 1 without the power. But in this case, the power makes the result approaches $ 0.05 $ instead.
Best Answer
From this step
$$\ldots=\lim_{n \to \infty} \left(1-\frac{3}{n}\right)^n =\ldots$$
we can't "plug" $\infty$ to solve since $1^\infty$ is an indeterminate form.
We can use
$$\left(1-\frac{3}{n}\right)^n=\left[\left(1+\frac{(-3)}{n}\right)^{\frac n{(-3)}}\right]^{-3}$$
and conclude by the standard limit
$$\lim_{x \to \pm\infty} \left(1+\frac{1}{x}\right)^x=e $$