In evaluating
$$\lim_{n\to \infty} \frac{(n-1)^n}{n^n}$$
I mechanically carried through the limit operator
\begin{align}\lim_{n\to \infty} \frac{(n-1)^n}{n^n}&= \lim_{n\to \infty} \Big(\frac{n-1}{n}\Big)^n\\&= \exp\left(\lim\limits_{n \to \infty}n\log\left(\frac{n-1}{n}\right)\right)\\&= \exp\left(\lim\limits_{n \to \infty}\frac{\log\left(\frac{n-1}{n}\right)}{\frac{1}{n}}\right)\\&= \exp\left(\lim\limits_{n \to \infty}\frac{\frac{1}{n^2 - n}}{\frac{-1}{n^2}}\right)\\&= \exp\left(\lim\limits_{n \to \infty}\frac{-n^2}{n^2 -n}\right)\\&= \exp(-1)\\&= \frac{1}{e} \end{align}
even though I didn't know that the limit exists. A user on this site told me this was bad form and that I shouldn't do this. However, I have always analyzed limits by assuming that the limit exists.
Suppose we are given the following limit for a real valued function $f(x)$
$$\lim_{x\to\infty}f(x)$$
where after evaluating this limit we find two different possibilities
- The limit converges to a finite number or functional value.
- The limit diverges to plus or minus infinity.
When evaluating this limit, can we always assume that the limit exists? We would therefore perform mechanical calculations and move the limit operator to reach one of these two possibilities.
What is so "bad" about carrying through the limit operator when we don't know that the limit exists? Is there an example where something goes horribly wrong? I don't remember reading about a specific theorem which states that you should not do this. Am I overlooking something?