Question

Limit question

Answer 1

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.

Answer 2

Any ideas?

Answer 3

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.

Answer 4

Hmm... Still trying...

Answer 5

Bah I can't get this!!

Answer 6

Wow. i got it. I am dumb.

Answer 7

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 \((fg)(x)=1\)

Answer 8

Or, if they can be the same function, define \(f\) and \(g\) as follows:

\[f(x)=g(x) =\begin{cases}
1 & x\in \mathbb{Q} \\
-1& x\notin \mathbb{Q}
\end{cases}
\]

Answer 9

What's Q?

Answer 10

The set of all rational numbers.

Answer 11

I got another one.

Answer 12

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

Answer 13

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

Answer 14

Cool! also explain why you used \(2\) instead of \(1\) as it was clever

Answer 15

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.

Answer 16

Right, I was just thinking to avoid dividing by 0

Answer 17

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

Answer 18

@zzr0ck3r .

Answer 19

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

Answer 20

Yes I have taken complex analysis haha but okay cool!

Answer 21

not complex, extended reals.

Answer 22

\([-\infty, \infty]\) is a thing and \(\infty\in [-\infty, \infty]\)

Answer 23

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