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mukushla
Group Title
Evaluate\[\lim_{x \rightarrow 0}x\lfloor \frac{1}{x} \rfloor\]
 one year ago
 one year ago
mukushla Group Title
Evaluate\[\lim_{x \rightarrow 0}x\lfloor \frac{1}{x} \rfloor\]
 one year ago
 one year ago

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MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
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 :/
 one year ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
I am taking Cal 1 too single variable. I just went over that, :) ( ironically I was struggling with a question like that yesterday)
 one year ago

datanewb Group TitleBest ResponseYou've already chosen the best response.0
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 \[\lim_{x\rightarrow \infty}x\lfloor \frac{1}{x} \rfloor\]
 one year ago

MarcLeclair Group TitleBest ResponseYou've already chosen the best response.0
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 ?
 one year ago

hellow Group TitleBest ResponseYou've already chosen the best response.1
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?
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
yes its 1
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
Hint : for all \(x\in \mathbb{R}\)\[0\le x\lfloor x \rfloor<1\]
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
it has to do something with squeeze theorem
 one year ago

hellow Group TitleBest ResponseYou've already chosen the best response.1
I would probably try an epsilondelta 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 epsilondelta 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!
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
thank u for ur effort :)
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
\[0\le \frac{1}{x}\lfloor \frac{1}{x} \rfloor<1\]\[\frac{1}{x}\le \lfloor \frac{1}{x} \rfloor<1\frac{1}{x}\]we're doin limit so x will not reach 0 so multiply both sides of later thing by x\[1\le x\lfloor \frac{1}{x} \rfloor<x1\]multiply by 1\[1x< x\lfloor \frac{1}{x} \rfloor\le 1\]apply limit\[\lim_{x \rightarrow 0}(1x)< \lim_{x \rightarrow 0}(x\lfloor \frac{1}{x} \rfloor)\le \lim_{x \rightarrow 0}1\]\[1< \lim_{x \rightarrow 0}(x\lfloor \frac{1}{x} \rfloor)\le 1\]so\[\lim_{x \rightarrow 0}x\lfloor \frac{1}{x} \rfloor=1\]
 one year ago

hellow Group TitleBest ResponseYou've already chosen the best response.1
Nice! I think that works, and am glad to see how to do it:).
 one year ago
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