0
$\begingroup$

$\displaystyle \lim_{x\rightarrow \infty}\left(x-\lfloor x\rfloor\right) = $

I have Tried like this way:: Let $x = k\rightarrow \infty$. Then Limit Convert into $\lim_{x\rightarrow k}(x-\lfloor x \rfloor)$

Now When $x = k\rightarrow \infty$. Then $x=\lfloor x \rfloor $.

So $\displaystyle \lim_{x\rightarrow \infty}\left(x-\lfloor x\rfloor\right) = 0$

But Answer given is Limit does not exists.

$\endgroup$
  • $\begingroup$ For what $N$ is it the case that $x > N$ implies the expression is $0$? How about $N = 1000$? No, that can't work, because if $x = 1000.5 > 1000$, then the expression is $0.5 \neq 0$... $\endgroup$ – Benjamin Dickman Oct 8 '13 at 4:11
  • $\begingroup$ sawtooth wave !!!. $\endgroup$ – Felix Marin Oct 8 '13 at 4:18
  • $\begingroup$ Benjamin Dickman ,Felix Marin $\endgroup$ – juantheron Oct 8 '13 at 8:25
5
$\begingroup$

The limit does not exist. Note that $x\mapsto x-\lfloor x\rfloor$ traces out all numbers in $[0,1)$ indefinitely many times as $x\to+\infty$ (to convince yourself, you may want to draw a graph; it's gonna be a sequence of segments pointing to the northeast). That is, there is not a particular number to which this function gets ever closer as $x$ tends to infinity (which, intuitively, is the very essence of the definition of convergence).

In your example, what you really show is that $$\lim_{\substack{x\to+\infty\\\text{$x$ is an integer}}}\left(x-\lfloor x\rfloor\right)=0.$$ This is elusive, because you have chosen only a particular sequence of $x$'s (namely, the sequence of integers) and shown that the associated sequence of function values converges to zero. However, by the definition of convergence, you ought to show this result for all sequences of $x$'s tending to infinity. But then again, you can't show this, because convergence does not occur.

$\endgroup$
  • 1
    $\begingroup$ Thanks triple_sec................. $\endgroup$ – juantheron Oct 8 '13 at 8:25
  • $\begingroup$ @juantheron My pleasure! $\endgroup$ – triple_sec Oct 8 '13 at 15:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.