Looking for something else?

Not the answer you are looking for? Search for more explanations.

- anonymous

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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

Get your **free** account and access **expert** answers to this and **thousands** of other questions

- anonymous

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

- chestercat

See more answers at brainly.com

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

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 :/

- anonymous

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

- datanewb

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

- anonymous

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 ?

- anonymous

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?

- anonymous

yes its 1

- anonymous

Hint :
for all \(x\in \mathbb{R}\)\[0\le x-\lfloor x \rfloor<1\]

- anonymous

it has to do something with squeeze theorem

- anonymous

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!

- anonymous

thank u for ur effort :)

- anonymous

\[0\le \frac{1}{x}-\lfloor \frac{1}{x} \rfloor<1\]\[-\frac{1}{x}\le -\lfloor \frac{1}{x} \rfloor<1-\frac{1}{x}\]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

- anonymous

Nice! I think that works, and am glad to see how to do it:).

Looking for something else?

Not the answer you are looking for? Search for more explanations.