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.
Best Answer
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.