Can we carry through the limit operator when we are unsure if the limit exists?

Question

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

1. The limit converges to a finite number or functional value.
2. 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?

Answer 1

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.

Answer 2

There is nothing wrong in your steps. However, there is no need to use L'Hospital's rule. You can evaluate it directly: $$\lim_{n\to \infty} \frac{(n-1)^n}{n^n}=\lim_{n\to \infty} \left[\left(1+\frac{1}{-n}\right)^{-n}\right]^{-1}= \left[\lim_{n\to \infty}\left(1+\frac{1}{-n}\right)^{-n}\right]^{-1}=e^{-1}.$$ Note: $(1+1/n)^n\to e$ from below, while $[1+1/(-n)]^{-n}\to e$ from above.

In the algebraic limit theorem it is stated:

taking the limit is compatible with the algebraic operations, provided the limits on the right sides of the equations below exist

So, as long as the limits on the right sides of your equations exist, it is an allowed operation.

Answer 3

One does not assume beforehand that a limit exists. While evaluating the limit of an expression you can proceed step by step following limit rules and if you succeed at the end then that is a proof that the limit exists and your approach is correct (there may be circularity involved depending upon what are the known facts but that's another issue).

Also the limit rules used in each step should be reversible or they need to applied correctly. Thus for example using L'Hospital's Rule if you succeed in finding a limit then it's fine. But if L'Hospital's Rule fails then one can't conclude that the original limit does not exist. The reversible versions of limit rules are explained in this answer (and their proof is discussed in this thread).

Another step which you have used in your solution is the interchange of a functional operation and limit operation. A sufficient condition for this to work is that the function involved be continuous which is the case here with $\exp$ being continuous.