Limits that take on a range rather than a unique value

Question

I've come across two limits which are reported to take on a range of values rather than a unique one.

They are:

$$\lim \limits_{x \to \infty} \space\frac{1+\cos x}{1-\sin x} = 0 \space to\space \infty$$

$$\lim \limits_{x \to \infty} \space \frac{2+2x+\sin(2x)}{(2x+\sin(2x))e^{\sin x}} = \frac{1}{e} \space to \space e$$

These answers seem to contradict what I know and understand as the definition of a limit. I'm self-studying maths so am clearly missing something here.

My questions:

1. How are these limits solved?
2. Why do they exist? Why is a range of values allowed here?
3. What kind of books should I read to learn more about this? If you could provide references that would be perfect.

ETA results from Wolfram:

The First The Second

Answer 1

In the usual sense (and common definition), these limits do not exist (since, loosely speaking, the function values don't come arbitrarily close to a specific value for sufficiently large values of $x$).

For a simpler example, consider: $$\lim_{x \to +\infty} \sin x$$ The fact that $\sin x$ keeps taking (all) values in the interval $[-1,1]$, even for arbitrary large values of $x$, can be used as an argument against the existence of this limit.

Of course, saying that $\sin x$ keeps taking values in $[-1,1]$ contains more information than simply saying "the limit does not exist", so it can be useful to consider (and determine) this range - although I would never say "the limit is [a range]".

Answer 2

The limits definitely do not exist for these expressions. Don't let any resources on the web convince you otherwise. However, that said, the limits inferior and superior may exist. So, a more accurate rendition of your first "limit" is: $$ \liminf_{x\to\infty}\frac{1+\cos(x)}{1-\sin(x)} = \lim_{x\to\infty}\inf_{t \ge x}\frac{1+\cos(t)}{1-\sin(t)} = 0 \text{ (as } \cos(t)=-1 \text{ infinitely often above any } x \text{)} $$ and $$ \limsup_{x\to\infty}\frac{1+\cos(x)}{1-\sin(x)} = \lim_{x\to\infty}\sup_{t \ge x}\frac{1+\cos(t)}{1-\sin(t)} = \infty \text{ (as } \sin(t)=1 \text{ infinitely often above any } x \text{)} $$

This is why the resources across the web claim that the "limit exists as a range of values between zero and infinity"

Answer 3

Take first limit for example it means that you can find an increasing sequence (because $x$ goes to $\infty$, in general the sequence doesn't need to be increasing it should just converge to the what $x$ is tending to). $x_n$ for any value of $L\in[0,\infty]$ so that $$\lim_{x\to\infty}\frac{1+\cos x_n}{1-\sin x_n}=L$$ The limit doesn't exist but in certain scenarios it's good to know what are the possible values in case you're looking for a particular sequence of $x$'s, for example $$\lim_{x\to \infty}x-\lfloor x\rfloor$$ Can be anything from $0$ to $1$ but if you're interested only in integers then the limit converges to $0$.