I was told that I couldn't "pull the limit in". Tell me exactly how I'm messing up, please!

Question

So, the problem that we were solving was

$$\lim_{n\to\infty}\left(\frac{n}{n+1}\right)^n$$

To figure out whether the series converged or diverged, after simplification, I asked my professor whether finding the limit of the inside of the function to determine whether the inside function was divergent or convergent would be helpful. My logic was this: If it's divergent, infinity to the power of $n$ as $n$ approaches infinity is just infinity, and if it's between $-1$ and $1$ then it approaches a finite value, right? Meaning that, if we were to take the limit of the inside it would ultimately determine what the function did after taken to the power of n.

I was told that we simply couldn't do this but her explanation was a bit lackluster, it was basically "because I said so." Please tell me exactly how I'm wrong so I can better understand what I'm doing. Thank you!

Answer 1

In general we can't pull a limit past a variable that occurs in the limiting operation, because that might change the result. Here's an example which is easier than the problem you're working on:

$$ \lim_{x \to 1}\ (x-1)^{x-1} $$

If we take the limit of the inside function first, we'll get $0$, but in fact, the correct answer is $1$.

Now it is true that you can slide a limit operation past a variable that doesn't occur in the limit:

$$ \lim_{x \to 0} (x^y) = (\lim_{x \to 0} x)^y = 0^y = 0. $$

And it's true that you can slide a limit operation past a continuous function:

$$ \lim_{x \to \infty} \log(1 + \frac{1}{x}) = \log\left(\lim_{x \to \infty} 1 + \frac{1}{x}\right) = \log(1) = 0. $$

So it's easy to get tempted to think you can slide limits past whatever you want, but in fact you'll get the wrong answer.

Answer 2

Explaining to someone "exactly why" they are wrong is actually quite difficult, because often their mistake is a result of fuzzy/intuitive reasoning and so its hard to pin down "exactly" the mistake. You say "my logic is..." but this isn't rigorous logic in the mathematical sense; you just aren't using the definition of the limit correctly.

1. Infinity is not a number. There is no such thing as "infinity to the $n$".
2. The limit of a sequence $a_n$ can never be given by an expression that depends on $n$. This wouldn't make any sense.
3. In general operations of this kind can be seen to be invalid by considering basic examples like e.g. $$ 1 = n/n = n \times \tfrac{1}{n}. $$ First take limit of $1/n$. Its zero. So you have $n \times 0$. And this is always zero, so the total limit is zero. OR how about first take limit of $n$ to get infinity. But now you have infinity times $1/n$ which is always infinity!?

4.You are essentially considering $$ b_{m,n} = \Bigl( \frac{m}{m+1}\Bigr)^n, $$ and saying "let's imagine taking the limit as $m \to \infty$ first and then taking the limit as $n \to \infty$. This sort of limit swapping is a classic issue in analysis that sometimes can be done and other times cannot.

Answer 3

I'm not entirely certain how you're arguing this, but I think you're observing that $$-1 \le \frac{n}{n+1} \le 1,$$ and so raising this sequence to the power of $n$, you have $$0 \le \left(\frac{n}{n+1}\right)^n \le 1^n = 1,$$ which you claim makes the limit of the sequence finite.

It almost works; the problem is that divergence to infinity isn't the only way a sequence can fail to converge. Another example is the sequence $\cos(\pi n / 2)$, which takes the values $1, 0, -1, 0, 1, 0, -1, 0, \ldots$. It fails to converge, but is still bounded.

If you want to show that a bounded sequence still has a limit, you can try using the monotone convergence theorem. In this case, with a bit of work, you can show the sequence is monotone decreasing, and therefore has a limit.

Answer 4

The reason why you can't do this is that the argument you've applied can lead you to the wrong conclusion, even for determining whether the limit is finite or not. For example, in the following limit the "inside" thing tends to $1$: $$ \lim_{n\to\infty}\left(1+\frac1{\sqrt n}\right)^n $$ so you would conclude that the limit is finite. It turns out that the limit is infinity.