Question Number 203822 by 073 last updated on 29/Jan/24
Answered by esmaeil last updated on 29/Jan/24
![1:(1/x)=p→^(x→∞→p→0) [((x+1)/x)]=[p+1)=1→ Ω=lim_(p→0) ((sinp)/p)=1](https://www.tinkutara.com/question/Q203823.png)
$$\mathrm{1}:\frac{\mathrm{1}}{{x}}={p}\overset{{x}\rightarrow\infty\rightarrow{p}\rightarrow\mathrm{0}} {\rightarrow}\left[\frac{{x}+\mathrm{1}}{{x}}\right]=\left[{p}+\mathrm{1}\right)=\mathrm{1}\rightarrow \\ $$$$\Omega=\underset{{p}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{sinp}}{{p}}=\mathrm{1} \\ $$
Answered by esmaeil last updated on 29/Jan/24
![2:Ω=lim_(x→1^− ) ((2x(1−x))/(1−x))=2 3:Ω=lim_(x→0^+ ) (([x]−x)/x)=lim_(x→0) ((0−x)/x)=−1](https://www.tinkutara.com/question/Q203824.png)
$$\mathrm{2}:\Omega=\underset{{x}\rightarrow\overset{−} {\mathrm{1}}} {\mathrm{lim}}\frac{\mathrm{2}{x}\left(\mathrm{1}−{x}\right)}{\mathrm{1}−{x}}=\mathrm{2} \\ $$$$\mathrm{3}:\Omega=\underset{{x}\rightarrow\overset{+} {\mathrm{0}}} {\mathrm{lim}}\frac{\left[{x}\right]−{x}}{{x}}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{0}−{x}}{{x}}=−\mathrm{1} \\ $$
Answered by MathematicalUser2357 last updated on 30/Jan/24
![What is: [[x]]](https://www.tinkutara.com/question/Q203850.png)
$${What}\:{is}:\:\left[\left[{x}\right]\right] \\ $$