Evaluating limits of floor functions

Question

I have 2 limits to calculate that I'm not sure if I've "proven" correctly.

(a) $\lim_{x\to 0}$ $x\lfloor{x}\rfloor$.

My guess is that since this has to be continuous, (however not sure if I have to prove this), I can just plug in 0 and I get

$\lim_{x\to 0}$ $x\lfloor{x}\rfloor$ $=$ $0\lfloor{0}\rfloor$ $=$ $0$

How can this be shown without just trying $0$?

(b) $\lim_{x\to \infty}$ $\frac{\lfloor{x}\rfloor}{x}$

My guess is that since $\lfloor{x}\rfloor$ is defined as the biggest number $\leq$ $x$, the limit has to be $1$. But I'm not quite sure what to prove here since my answer is merely a guess.

How do approach this? Is there a simple way to calculate these?

Answer 1

1. For $h\to0^+,\lfloor h\rfloor=0$ so the RHL is $0$. For $h\to0^-,\lfloor h\rfloor=-1$ so LHL$=\lim_{h\to0^-}-h=0$. So the limit is $0$. You can't say that $f(x)$ is continuous at $0$ and just substitute $0$ to get the limit since that is what you are supposed to show here.

2. Note that $\lfloor x\rfloor=x-\{x\}$ where $\{x\}\in[0,1)$ is the fractional part of $x$. The limit becomes$$\lim_{x\to\infty}\frac{x-\{x\}}x=1-0=1.$$

Answer 2

(a) The left-hand limit is that of $-x$, since $\lfloor x\rfloor=-1$ in $[-1,0)$. The right-hand limit is that of $0$, since $\lfloor x\rfloor=0$ in $[0,1)$. The two one-sided limits agree at $0$, so the ordinary limit is also $0$.

(b) Rewrite this as $\frac{x-\{x\}}x=1-\frac{\{x\}}x$, where $\{x\}$ is the fractional part of $x$ (always in $[0,1)$). $\frac{\{x\}}x$ can be squeezed to $0$, so the limit is $1$.