A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
Limit question
anonymous
 one year ago
Limit question

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hmm... Still trying...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Bah I can't get this!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Wow. i got it. I am dumb.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2Define \(f\) 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 \((fg)(x)=1\)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2Or, if they can be the same function, define \(f\) and \(g\) as follows: \[f(x)=g(x) =\begin{cases} 1 & x\in \mathbb{Q} \\ 1& x\notin \mathbb{Q} \end{cases} \]

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2The set of all rational numbers.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0cos(x)+2 and g(x)=1/(cos(x)+2)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Both don't exist but multiplied, it's 1.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2Cool! also explain why you used \(2\) instead of \(1\) as it was clever

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Cuz 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.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2Right, I was just thinking to avoid dividing by 0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0One question for you. Does an infinite limit imply non existent limit?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2yes, since we are in the real numbers. But in the extended real numbers, it is a thing !

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes I have taken complex analysis haha but okay cool!

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2not complex, extended reals.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2\([\infty, \infty]\) is a thing and \(\infty\in [\infty, \infty]\)

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.2But yes, there are definitely two types of not converging, one where the function has a converging subset, and one where it does not.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.