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?
Best Answer
It's a shorthand notation, but is not really "bad form" if both you and your readers understand what is really going on. What you really mean at each step where you write something like "$\lim_{n \to \infty} f(n) = \lim_{n \to \infty} g(n)$" is "if $\lim_{n \to \infty} g(n)$ exists, then $\lim_{n \to \infty} f(n)$ exists and has the same value". When at the end of the calculation you find that the last limit does exist, that tells you that everything is good and you have found the limit you wanted. If you ended up with a limit that doesn't exist, then you might not know about the original limit.
EDIT: In particular, I note that you are using l'Hopital's rule to go from $\lim_{n \to \infty} \log\left(\frac{n-1}{n}\right)/(1/n)$ to $\lim_{n \to \infty} (1/(n^2-n))/(1/n^2)$. This is OK since the rule says if $f(n)$ and $g(n)$ both go to $0$ or $\infty$ as $n \to \infty$ and $\lim_{n \to \infty} f'(n)/g'(n) = L$ exists, then $\lim_{n \to \infty} f(n)/g(n) = L$ as well.
You shouldn't use the rule in the other direction, since it can happen that $\lim_{n \to \infty} f(n)/g(n)$ exists but $\lim_{n \to \infty} f'(n)/g'(n)$ does not.