## anonymous one year ago Limit question

1. anonymous

2. anonymous

I can easily think of f(x) and g(x) that does not exist (1/x^2 and 1/x^4 to name a few...) but I can't think of anything that the limit of the product would give a 1.

3. anonymous

Any ideas?

4. anonymous

I can't think of an exact example, but as x approaches infinity the value will be 0. Cos (0) will equal 1. Does this help.

5. anonymous

Hmm... Still trying...

6. anonymous

Bah I can't get this!!

7. anonymous

Wow. i got it. I am dumb.

8. zzr0ck3r

Define $$f$$ and $$g$$ as follows: $f(x) =\begin{cases} \frac{1}{2} & x\in \mathbb{Q} \\ -\frac{1}{2} & x\notin \mathbb{Q} \end{cases}$ $g(x) =\begin{cases} 2 & x\in \mathbb{Q} \\ -2& x\notin \mathbb{Q} \end{cases}$ Then $$(fg)(x)=1$$

9. zzr0ck3r

Or, if they can be the same function, define $$f$$ and $$g$$ as follows: $f(x)=g(x) =\begin{cases} 1 & x\in \mathbb{Q} \\ -1& x\notin \mathbb{Q} \end{cases}$

10. anonymous

What's Q?

11. zzr0ck3r

The set of all rational numbers.

12. anonymous

I got another one.

13. anonymous

cos(x)+2 and g(x)=1/(cos(x)+2)

14. anonymous

Both don't exist but multiplied, it's 1.

15. zzr0ck3r

Cool! also explain why you used $$2$$ instead of $$1$$ as it was clever

16. anonymous

Cuz by squeeze theorem if we uses cos(x)+1 then we have 1/(-1+1) for the limit on the LHS limit which can't be determined.

17. zzr0ck3r

Right, I was just thinking to avoid dividing by 0

18. anonymous

One question for you. Does an infinite limit imply non existent limit?

19. anonymous

@zzr0ck3r .

20. zzr0ck3r

yes, since we are in the real numbers. But in the extended real numbers, it is a thing !

21. anonymous

Yes I have taken complex analysis haha but okay cool!

22. zzr0ck3r

not complex, extended reals.

23. zzr0ck3r

$$[-\infty, \infty]$$ is a thing and $$\infty\in [-\infty, \infty]$$

24. zzr0ck3r

But yes, there are definitely two types of not converging, one where the function has a converging subset, and one where it does not.