anonymous
  • anonymous
Limit question
Mathematics
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anonymous
  • anonymous
Limit question
Mathematics
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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anonymous
  • anonymous
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anonymous
  • 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.
anonymous
  • anonymous
Any ideas?

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anonymous
  • 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.
anonymous
  • anonymous
Hmm... Still trying...
anonymous
  • anonymous
Bah I can't get this!!
anonymous
  • anonymous
Wow. i got it. I am dumb.
zzr0ck3r
  • 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 \((fg)(x)=1\)
zzr0ck3r
  • zzr0ck3r
Or, if they can be the same function, define \(f\) and \(g\) as follows: \[f(x)=g(x) =\begin{cases} 1 & x\in \mathbb{Q} \\ -1& x\notin \mathbb{Q} \end{cases} \]
anonymous
  • anonymous
What's Q?
zzr0ck3r
  • zzr0ck3r
The set of all rational numbers.
anonymous
  • anonymous
I got another one.
anonymous
  • anonymous
cos(x)+2 and g(x)=1/(cos(x)+2)
anonymous
  • anonymous
Both don't exist but multiplied, it's 1.
zzr0ck3r
  • zzr0ck3r
Cool! also explain why you used \(2\) instead of \(1\) as it was clever
anonymous
  • 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.
zzr0ck3r
  • zzr0ck3r
Right, I was just thinking to avoid dividing by 0
anonymous
  • anonymous
One question for you. Does an infinite limit imply non existent limit?
anonymous
  • anonymous
zzr0ck3r
  • zzr0ck3r
yes, since we are in the real numbers. But in the extended real numbers, it is a thing !
anonymous
  • anonymous
Yes I have taken complex analysis haha but okay cool!
zzr0ck3r
  • zzr0ck3r
not complex, extended reals.
zzr0ck3r
  • zzr0ck3r
\([-\infty, \infty]\) is a thing and \(\infty\in [-\infty, \infty]\)
zzr0ck3r
  • 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.

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