We know that

$\lfloor x\rfloor$ : greatest integer $\leq x$

$\lceil x\rceil$ : least integer $\geq x$

$\mathrm{frac}(x)=x-\lfloor x\rfloor$

I can solve the questions limit of function like $$ \lim\limits_{x\to n^\pm}\frac{\lfloor x-1\rfloor}{x-1}\\ \lim\limits_{x\to n^\pm}\frac{\lfloor x\rfloor}{x-1} $$ As $x$ approaches $n$ from above, $\lfloor x-1\rfloor=n-1$; therefore, $$ \lim_{\large x\to n^+}\frac{\lfloor x-1\rfloor}{x-1}=\frac{n-1}{n-1} $$ but I can't solve the questions like

$$ \lim\limits_{x\to\infty}\frac{\lfloor x-3\rfloor}{x-1},\\ \lim\limits_{x\to\infty}\frac{\lceil x-3\rceil}{x-1},\\ \lim\limits_{x\to a} \lfloor x\rfloor,\\ \lim\limits_{x\to a} \lceil x\rceil,\\ \lim\limits_{x\to a} \mathrm{frac}( x) $$ for $a\in \mathbf{R}$

All questions are similar type, so I have given many problems in my questions. Please help me.

  • $\begingroup$ None of the limits possibly exist. $\endgroup$ Sep 1, 2016 at 10:51

1 Answer 1


For the limit at infinity:

$$L=\lim_{x\to \infty}\frac{\lfloor x-3\rfloor}{x-1}=\lim_{x\to \infty}\frac{x-3-\text{frac}(x-3)}{x-1}=\lim_{x\to \infty}\frac{x-3}{x-1}-\frac{\text{frac}(x-3)}{x-1}$$ since $$0\leq\frac{\text{frac}(x-3)}{x-1}<\frac{1}{x-1}$$ we have $$L=\lim_{x\to \infty}\frac{x-3}{x-1}=1$$

For the limit at a:

Let $\epsilon = \frac{\text{frac}(a)}{2}$, we have that $\lfloor x\rfloor=\lfloor a\rfloor$ for $x\in(a-\epsilon,a+\epsilon).$ So $\lim_{x\to a} \lfloor x\rfloor=\lfloor a\rfloor.$

NOTE THIS ASSUMES $\text{frac}(a) > 0$, if $a\in \mathbb{Z}$ then the limit does not exist.

  • $\begingroup$ what can we say about ceiling function in my questions $\endgroup$
    – user356595
    Sep 1, 2016 at 4:27

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