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anonymous

  • one year ago

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    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
    • one year ago
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    Any ideas?

  4. anonymous
    • one year ago
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    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
    • one year ago
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    Hmm... Still trying...

  6. anonymous
    • one year ago
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    Bah I can't get this!!

  7. anonymous
    • one year ago
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    Wow. i got it. I am dumb.

  8. zzr0ck3r
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    What's Q?

  11. zzr0ck3r
    • one year ago
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    The set of all rational numbers.

  12. anonymous
    • one year ago
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    I got another one.

  13. anonymous
    • one year ago
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    cos(x)+2 and g(x)=1/(cos(x)+2)

  14. anonymous
    • one year ago
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    Both don't exist but multiplied, it's 1.

  15. zzr0ck3r
    • one year ago
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    Cool! also explain why you used \(2\) instead of \(1\) as it was clever

  16. anonymous
    • one year ago
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    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
    • one year ago
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    Right, I was just thinking to avoid dividing by 0

  18. anonymous
    • one year ago
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    One question for you. Does an infinite limit imply non existent limit?

  19. anonymous
    • one year ago
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    @zzr0ck3r .

  20. zzr0ck3r
    • one year ago
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    yes, since we are in the real numbers. But in the extended real numbers, it is a thing !

  21. anonymous
    • one year ago
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    Yes I have taken complex analysis haha but okay cool!

  22. zzr0ck3r
    • one year ago
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    not complex, extended reals.

  23. zzr0ck3r
    • one year ago
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    \([-\infty, \infty]\) is a thing and \(\infty\in [-\infty, \infty]\)

  24. zzr0ck3r
    • one year ago
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    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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