Manan

Not entirely sure, but I think the product rule for limits doesn't extend to infinite number of multiplicands(I don't think infinite number of multiplicands make sense even in general for $\Bbb{R})$.

Just a moment ago I discovered the same question with a very mild change. Should I consider deleting my question? (Here's the link to that similar question: math.stackexchange.com/questions/508733/…)

I posted a question a few moments back, to which I've accepted an answer(the answer is enlightening, and already has 4 upvotes). Here's the link: math.stackexchange.com/questions/3817624/…

The question has been answered

@SpecterProphet Although I haven't studied analysis yet, this question has dumbfounded me. Just like the asker, I'm of the view that such a function does not exist, but all sorts of weird functions exist in higher math so I'm curious to know what the answer actually is.

I'd really like someone to answer this question: math.stackexchange.com/questions/3788377/…

@LucasHenrique It's as good as $^{\text{}}$

Thanks @AkivaWeinberger - I found on TeX SE $\textsuperscript{text goes here}$

I need help formatting plain text with spaces in an exponential. Can anyone help?

By clicking on the image profile here, you can create a new chat-room with a user.

Ah, I found out

:)

I guess the site suggests to move extended discussions to chat. For other chat-rooms(such as CURED and the likes), I'm not aware of the process.

You could add your calculations as a final answer which can be accepted.

I see. Then it most certainly is (1) Imprecise phrasing/solution in the book, or (2) Imprecise phrasing by the asker

@Axel Isn't that precisely the claim that the math book makes- that the number of 4 digit numbers such that their product of digits is divisible by 7 is 4904? I believe the number of 4 digit numbers, not including 0 as a digit, not divisible by 7 is 4904.

Ah, I see what you meant; so you're including all the $4$ digit numbers, even those including $0$ when you accounted $4904$.

@Axel I guess you meant not divisible by 7 in your last comment?

@CalvinKhor Thanks!

If a function $f:X\to X$ is defined such that $f\left(f(x)\right)=x$ $\forall x\in X$, is $f$ necessarily bijective? Intuitively I think it should be, but I can't prove this rigorously.