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Consider $f(x) = \lfloor x \rfloor + \lfloor -x \rfloor $ . Now find value of $\lim_{x \to \infty} f(x) $ . I know that if $x_0 \in \mathbb{R}$ then $\lim_{x \to x_0} f(x) = -1$ but I don't know whether it is true or not in the infinity .

S.H.W
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    $x\in \mathbb Z\implies f(x)=x+(-x)=0$. $x\notin \mathbb Z\implies f(x)=-1$ so the limit does not exist. – lulu May 06 '17 at 21:02
  • @lulu You are wrong . We are talking about limit not value of function . – S.H.W May 06 '17 at 21:03
  • lulu 's point is more or less the same as boaz 's answer, it boils down to the fact that the function do not converge because for arbitrarily large values, $f$ jumps between the values $0$ and $-1$. – N.Bach May 06 '17 at 21:07
  • No, I'm not wrong. $f(x)$ is a function which hits $0$ for arbitrarily large $x$ and it hits $-1$ for arbitrarily large $x$. That means it can not reach a limit at $\infty$. – lulu May 06 '17 at 21:14
  • @lulu Can you explain why this happens ? https://www.wolframalpha.com/input/?i=limit+floor(x)+%2B+floor(-x)+as+x+to+infinity – S.H.W May 06 '17 at 21:16
  • Sure. Wolfram Alpha made a mistake. – lulu May 06 '17 at 21:16

2 Answers2

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It has no limit when $x\to\infty$. Consider the sequences $$ x_n=n\qquad y_n=n+\frac{1}{2} $$ Both sequence tend to $\infty$, but notice that $f(x_n)=0$ while $f(y_n)=-1$ for every $n$.

boaz
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Defining with $ \left\{ x \right\}$ the fractional part of $x$, i.e. $$ x = \left\lfloor x \right\rfloor + \left\{ x \right\} $$ and denoting by $[P]$ the Iverson bracket $$ \left[ P \right] = \left\{ {\begin{array}{*{20}c} 1 & {P = TRUE} \\ 0 & {P = FALSE} \\ \end{array} } \right. $$ then we have that $$ \begin{array}{l} f(x) = \left\lfloor x \right\rfloor + \left\lfloor { - x} \right\rfloor = \left\lfloor x \right\rfloor - \left\lceil x \right\rceil = - \left( {\left\lceil x \right\rceil - \left\lfloor x \right\rfloor } \right) = \\ = - \left\lceil {\left\{ x \right\}} \right\rceil = - 1 + \left[ {x \in Z} \right] = \left\{ {\begin{array}{*{20}c} { - 1} & {\left| {\;x \notin Z} \right.} \\ 0 & {\left| {\;x \in Z} \right.} \\ \end{array}} \right. \\ \end{array} $$

and the limit for $x \to \infty$ does not exist.

G Cab
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