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mukushla

  • 3 years ago

Evaluate\[\lim_{x \rightarrow 0}x\lfloor \frac{1}{x} \rfloor\]

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  1. MarcLeclair
    • 3 years ago
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    It does not exist ax you can't factor anything out and will end up with 0/0.Which is an undetermined form. I hope I helped :/

  2. MarcLeclair
    • 3 years ago
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    I am taking Cal 1 too single variable. I just went over that, :) ( ironically I was struggling with a question like that yesterday)

  3. datanewb
    • 3 years ago
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    Well, the equation is not defined at x=0, but as x approaches zero, would it's limit not be 1? @MarcLeclair, do you take the same stance for \[\lim_{x\rightarrow \infty}x\lfloor \frac{1}{x} \rfloor\]

  4. MarcLeclair
    • 3 years ago
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    I think that limit would then be infinity ( positive infinity) . However, I do not know why this sounds wrong to me, it is really puzzling my brain. As for his equation, it is because you will ALWAYS end up 0/0 which Doest not exist. Unless there is another way to define it ?

  5. hellow
    • 3 years ago
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    As for the second question, we have \[ \lim_{x \rightarrow \infty } x \lfloor \frac{ 1 }{ x } \rfloor = 0\], because, as long as x>1, \[\lfloor \frac{ 1 }{ x } \rfloor = 0\] You can make x as large as you like and this will still be true. Therefore, the product \[x \lfloor \frac{ 1 }{ x } \rfloor\] must also be zero, no matter how large x becomes. In other words, the product is not getting closer and closer to 1, it is consistently zero, i.e. {0,0,0,...}. As for the first question, I think\[ \lim_{x \rightarrow 0} x \lfloor \frac{ 1 }{ x } \rfloor \] does not exist. For any x=1/n, the value will be 1. For instance \[1/4\lfloor \frac{ 1 }{ \frac{ 1 }{ 4 } } \rfloor = 1/4\lfloor 4 \rfloor = \frac{ 1 }{ 4 }4 = 1\]. However, as x gets closer to zero, is the limit approaching 1?

  6. mukushla
    • 3 years ago
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    yes its 1

  7. mukushla
    • 3 years ago
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    Hint : for all \(x\in \mathbb{R}\)\[0\le x-\lfloor x \rfloor<1\]

  8. mukushla
    • 3 years ago
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    it has to do something with squeeze theorem

  9. hellow
    • 3 years ago
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    I would probably try an epsilon-delta proof if you are comfortable with those. The squeeze theorem is probably easier, but I don't see right now how it would work. For an epsilon-delta proof, if you want your product to be less than some epsilon, choose some 1/n which is less than epsilon. Then choose x= <1/(n+1). That way you product could be n/(n+1), which would put you within a distance of 1/(n+1) from 1 (close enough). And if x is smaller than 1/(n+1), you would have to show that you would still be within a distance 1/n from 1. The details might be annoying, and there is probably a better way, I just don't see it!

  10. mukushla
    • 3 years ago
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    thank u for ur effort :)

  11. mukushla
    • 3 years ago
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    \[0\le \frac{1}{x}-\lfloor \frac{1}{x} \rfloor<1\]\[-\frac{1}{x}\le -\lfloor \frac{1}{x} \rfloor<1-\frac{1}{x}\]we're doin limit so x will not reach 0 so multiply both sides of later thing by x\[-1\le -x\lfloor \frac{1}{x} \rfloor<x-1\]multiply by -1\[1-x< x\lfloor \frac{1}{x} \rfloor\le 1\]apply limit\[\lim_{x \rightarrow 0}(1-x)< \lim_{x \rightarrow 0}(x\lfloor \frac{1}{x} \rfloor)\le \lim_{x \rightarrow 0}1\]\[1< \lim_{x \rightarrow 0}(x\lfloor \frac{1}{x} \rfloor)\le 1\]so\[\lim_{x \rightarrow 0}x\lfloor \frac{1}{x} \rfloor=1\]

  12. hellow
    • 3 years ago
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    Nice! I think that works, and am glad to see how to do it:).

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