Question: 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.

Answer: 
One of the many ways
$$a\_n=\left(\frac{n-3}{n}\right)^n=\left(1-\frac{3}{n}\right)^n\implies \log(a\_n)=n\log\left(1-\frac{3}{n}\right)$$ Now, using Taylor
$$\log(1+\epsilon)=\epsilon -\frac{\epsilon ^2}{2}+O\left(\epsilon ^3\right)$$
$$\log\left(1-\frac{3}{n}\right)=-\frac{3}{n}-\frac{9}{2 n^2}+O\left(\frac{1}{n^3}\right)$$
$$\log(a\_n)=-3-\frac{9}{2 n}+O\left(\frac{1}{n^2}\right)$$
$$a\_n=e^{\log(a\_n)}=\frac{1}{e^3}-\frac{9}{2 e^3 n}+O\left(\frac{1}{n^2}\right)$$ shows the limit and how it is approached.
As you proceed,

$$\lim\_{n \to \infty} (1-\frac{3}{n})^n 
=\lim\_{n \to \infty} e^{n \ln(1-\frac{3}{n})}
 =\lim\_{n \to \infty} e^{\frac{ \ln(1-\frac{3}{n})}{1/n}} =e^{\lim\_{n \to \infty} \frac{\frac{3}{n^2}}{\frac{-1}{n^2} (1-\frac{3}{n})}} ( \text{by L'hospital rule}) $$

So, $$\lim\_{n \to \infty} (1-\frac{3}{n})^n = e^{-3} $$