Question Number 201657 by LimPorly last updated on 10/Dec/23
![if f(x)= { ((((sin (1+[x]))/([x])) for [x]≠0)),((0 for [x]=0)) :} where [x] represents an integer x greatest ≤ x Find lim_(x→0^− ) f(x).](https://www.tinkutara.com/question/Q201657.png)
$${if}\:\:{f}\left({x}\right)=\begin{cases}{\frac{\mathrm{sin}\:\left(\mathrm{1}+\left[{x}\right]\right)}{\left[{x}\right]}\:\:{for}\:\left[{x}\right]\neq\mathrm{0}}\\{\mathrm{0}\:\:{for}\:\left[{x}\right]=\mathrm{0}}\end{cases} \\ $$$${where}\:\left[{x}\right]\:{represents}\:{an}\:{integer}\:\boldsymbol{{x}}\:{greatest}\:\leqslant\:\boldsymbol{{x}} \\ $$$${Find}\:\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}{f}\left({x}\right). \\ $$
Answered by aleks041103 last updated on 10/Dec/23
![If I understand correctly: [x]=max{n:n∈Z ∧ x≥n} i.e. [2]=2; [3.5]=3; [−1.5]=−2 and so on. Then lim_(x→0^− ) [x]=lim_(ε→0^+ ) [0−ε]=lim_(ε→0^+ ) [−ε]=−1 ⇒lim_(x→0^− ) f(x)=( { ((((sin (1+[x]))/([x])) for [x]≠0)),((0 for [x]=0)) :})_([x]=−1) = =((sin(1+(−1)))/((−1)))=((sin(0))/(−1))=0](https://www.tinkutara.com/question/Q201658.png)
$${If}\:{I}\:{understand}\:{correctly}: \\ $$$$\left[{x}\right]={max}\left\{{n}:{n}\in\mathbb{Z}\:\wedge\:{x}\geqslant{n}\right\} \\ $$$${i}.{e}. \\ $$$$\left[\mathrm{2}\right]=\mathrm{2};\:\left[\mathrm{3}.\mathrm{5}\right]=\mathrm{3};\:\left[−\mathrm{1}.\mathrm{5}\right]=−\mathrm{2}\:{and}\:{so}\:{on}. \\ $$$$ \\ $$$${Then} \\ $$$$\underset{{x}\rightarrow\mathrm{0}^{−} } {{lim}}\left[{x}\right]=\underset{\varepsilon\rightarrow\mathrm{0}^{+} } {{lim}}\:\left[\mathrm{0}−\varepsilon\right]=\underset{\varepsilon\rightarrow\mathrm{0}^{+} } {{lim}}\left[−\varepsilon\right]=−\mathrm{1} \\ $$$$\Rightarrow\underset{{x}\rightarrow\mathrm{0}^{−} } {{lim}}\:{f}\left({x}\right)=\left(\begin{cases}{\frac{\mathrm{sin}\:\left(\mathrm{1}+\left[{x}\right]\right)}{\left[{x}\right]}\:\:{for}\:\left[{x}\right]\neq\mathrm{0}}\\{\mathrm{0}\:\:{for}\:\left[{x}\right]=\mathrm{0}}\end{cases}\right)_{\left[{x}\right]=−\mathrm{1}} = \\ $$$$=\frac{{sin}\left(\mathrm{1}+\left(−\mathrm{1}\right)\right)}{\left(−\mathrm{1}\right)}=\frac{{sin}\left(\mathrm{0}\right)}{−\mathrm{1}}=\mathrm{0} \\ $$
Commented by LimPorly last updated on 10/Dec/23
$${Thank}\:{you}\:{sir} \\ $$