Question Number 72806 by Rio Michael last updated on 03/Nov/19
![show that lim_( x→0) [ x] does not exist. Hence define [x] and sketch a graph for y = 3x^2 + [x]](https://www.tinkutara.com/question/Q72806.png)
$$\underset{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}\rightarrow\mathrm{0}} {\:{show}\:{that}\:\:\mathrm{lim}}\:\left[\:{x}\right]\:\:{does}\:{not}\:{exist}. \\ $$$${Hence}\:{define}\:\:\left[{x}\right]\:\:{and}\:{sketch}\:{a}\:{graph}\:{for}\: \\ $$$$\:{y}\:=\:\mathrm{3}{x}^{\mathrm{2}} \:+\:\left[{x}\right] \\ $$
Commented by mathmax by abdo last updated on 03/Nov/19
![let f(x)=[x] we have lim_(x→0^+ ) f(x)=lim_(x→0^+ ) [x] =0 lim_(x→0^− ) f(x)=lim_(x→0^− ) [x]=−1 and f(0)=0 we have 0≠−1 ⇒ lim_(x→0) [x] dont exist](https://www.tinkutara.com/question/Q72821.png)
$${let}\:{f}\left({x}\right)=\left[{x}\right]\:{we}\:{have}\:{lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:{f}\left({x}\right)={lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:\left[{x}\right]\:=\mathrm{0} \\ $$$${lim}_{{x}\rightarrow\mathrm{0}^{−} } \:\:{f}\left({x}\right)={lim}_{{x}\rightarrow\mathrm{0}^{−} } \:\:\:\left[{x}\right]=−\mathrm{1}\:\:{and}\:{f}\left(\mathrm{0}\right)=\mathrm{0}\:{we}\:{have}\:\mathrm{0}\neq−\mathrm{1}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \left[{x}\right]\:{dont}\:{exist} \\ $$
Commented by Rio Michael last updated on 03/Nov/19
$${love}\:{that}\:{method},\:{thanks}\:{sir} \\ $$
Answered by mind is power last updated on 03/Nov/19
![suppose that lim_(x→0) [x] exist call l=lim_(x→0) [x] ⇒∀ε>0 ∃η>0∣ ∣x∣<η⇒∣[x]−l∣<ε let ε=(l/3) ⇒∃η_ε such that ∀x∈]−η,η[ ∣[x]−L∣<(1/3) for x∈]−η,0[ ⇒∣−1−L∣<(1/2)⇔∣1+L∣<(1/3) for x∈]0,η[⇒∣−L∣<(1/3) we have ∣1+L−L∣=1 truangular inquality⇒∣1∣=1≤∣−L∣+∣1+L∣≤(1/3)+(1/3)=(2/3) absurd since that L/dosent exist 2) y=3x^2 +k x∈[k,k+1[](https://www.tinkutara.com/question/Q72812.png)
$$\mathrm{suppose}\:\mathrm{that}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\mathrm{x}\right]\:\mathrm{exist} \\ $$$$\mathrm{call}\:\mathrm{l}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\mathrm{x}\right] \\ $$$$\Rightarrow\forall\varepsilon>\mathrm{0}\:\exists\eta>\mathrm{0}\mid\:\:\:\mid\mathrm{x}\mid<\eta\Rightarrow\mid\left[\mathrm{x}\right]−\mathrm{l}\mid<\varepsilon \\ $$$$\mathrm{let}\:\epsilon=\frac{\mathrm{l}}{\mathrm{3}} \\ $$$$\left.\Rightarrow\exists\eta_{\varepsilon} \:\mathrm{such}\:\mathrm{that}\:\forall\mathrm{x}\in\right]−\eta,\eta\left[\:\:\mid\left[\mathrm{x}\right]−\mathrm{L}\mid<\frac{\mathrm{1}}{\mathrm{3}}\right. \\ $$$$\left.\mathrm{for}\:\mathrm{x}\in\right]−\eta,\mathrm{0}\left[\:\Rightarrow\mid−\mathrm{1}−\mathrm{L}\mid<\frac{\mathrm{1}}{\mathrm{2}}\Leftrightarrow\mid\mathrm{1}+\mathrm{L}\mid<\frac{\mathrm{1}}{\mathrm{3}}\right. \\ $$$$\left.\mathrm{for}\:\mathrm{x}\in\right]\mathrm{0},\eta\left[\Rightarrow\mid−\mathrm{L}\mid<\frac{\mathrm{1}}{\mathrm{3}}\right. \\ $$$$\mathrm{we}\:\mathrm{have}\:\mid\mathrm{1}+\mathrm{L}−\mathrm{L}\mid=\mathrm{1} \\ $$$$\mathrm{truangular}\:\mathrm{inquality}\Rightarrow\mid\mathrm{1}\mid=\mathrm{1}\leqslant\mid−\mathrm{L}\mid+\mid\mathrm{1}+\mathrm{L}\mid\leqslant\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\mathrm{absurd}\:\mathrm{since}\:\mathrm{that}\:\mathrm{L}/\mathrm{dosent}\:\mathrm{exist} \\ $$$$ \\ $$$$\left.\mathrm{2}\right)\: \\ $$$$\mathrm{y}=\mathrm{3x}^{\mathrm{2}} +\mathrm{k}\:\:\mathrm{x}\in\left[\mathrm{k},\mathrm{k}+\mathrm{1}\left[\right.\right. \\ $$
Commented by Rio Michael last updated on 03/Nov/19
$${thanks} \\ $$
Commented by mind is power last updated on 03/Nov/19
$$\mathrm{most}\:\mathrm{Welcom} \\ $$