Floor + Ceiling

Let $f(x) = \left \lfloor x \right \rfloor + \left \lceil x \right \rceil$ . We need to find $\displaystyle \lim _{x \to 0 } f(|x|)$ .

Note that $f(|x|)$ is an even function, because $|x| = |-x|$ for all $x$ , which implies $f(|x|) = f(|-x|)$ for all $x$ . This means that $f(|x|)$ is symmetric about the line $x = 0$ .

Hence we can conclude that $\displaystyle \lim _{x \to 0^{+}} f(|x|) = \displaystyle \lim _{x \to 0^{-}} f(|x|)$ .

Since both the one-sided limits are equal, the function $f(|x|)$ takes on finite values, and doesn't oscillate very quickly, we can say that the limit exists. We will now find its value.

$\begin{aligned} \lim_{x \to 0} f(|x|) & = \lim_{x \to 0^{+}} f(|x|) \\ & = \lim_{x \to 0^{+}} f(x) \\ & = \lim_{x \to 0^{+}} \left \lfloor x \right \rfloor + \left \lceil x \right \rceil \\ & = \lim_{x \to 0^{+}} \left \lfloor x \right \rfloor + \lim_{x \to 0^{+}}\left \lceil x \right \rceil \\ & = 0 + 1 = \boxed{1} & _\square \end{aligned}$

$\begin{aligned} \text{Left Hand Limit } & = \displaystyle \lim _{ x\to { 0 }^{ - } } \left\lfloor |x| \right\rfloor +\left\lceil |x| \right\rceil \\ & = \lim _{ h\to { 0 } } \ \left\lfloor |-h| \right\rfloor +\left\lceil |-h| \right\rceil \\ & = \lim _{ h\to { 0 } } \ \left\lfloor h \right\rfloor +\left\lceil h \right\rceil \\ & = 0 + 1 \\ & =1 \end{aligned}$

$\begin{aligned} \text{Right Hand Limit } & = \lim _{ x\to { 0 }^{ + } } \left\lfloor |x| \right\rfloor +\left\lceil |x| \right\rceil \\ & =\lim _{ h\to { 0 } } \left\lfloor |h| \right\rfloor +\left\lceil |h| \right\rceil \\ & = \lim _{ h\to { 0 } } \left\lfloor h \right\rfloor +\left\lceil h \right\rceil \\ & = 0 + 1 \\ & =1 \end{aligned}$

LHL = RHL =1

So, $\displaystyle \lim _{ x\to { 0 } } \left\lfloor |x| \right\rfloor +\left\lceil |x| \right\rceil =1.$