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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]

$$\underset{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}\rightarrow\mathrm{0}} {\:{show}\:{that}\:\:\mathrm{lim}}\:\left[\:{x}\right]\:\:{does}\:{not}\:{exist}. \\ $$$${Hence}\:{define}\:\:\left[{x}\right]\:\:{and}\:{sketch}\:{a}\:{graph}\:{for}\: \\ $$$$\:{y}\:=\:\mathrm{3}{x}^{\mathrm{2}} \:+\:\left[{x}\right] \\ $$

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

$${let}\:{f}\left({x}\right)=\left[{x}\right]\:{we}\:{have}\:{lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:{f}\left({x}\right)={lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:\left[{x}\right]\:=\mathrm{0} \\ $$$${lim}_{{x}\rightarrow\mathrm{0}^{−} } \:\:{f}\left({x}\right)={lim}_{{x}\rightarrow\mathrm{0}^{−} } \:\:\:\left[{x}\right]=−\mathrm{1}\:\:{and}\:{f}\left(\mathrm{0}\right)=\mathrm{0}\:{we}\:{have}\:\mathrm{0}\neq−\mathrm{1}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \left[{x}\right]\:{dont}\:{exist} \\ $$

Commented by Rio Michael last updated on 03/Nov/19

love that method, thanks sir

$${love}\:{that}\:{method},\:{thanks}\:{sir} \\ $$

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[

$$\mathrm{suppose}\:\mathrm{that}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\mathrm{x}\right]\:\mathrm{exist} \\ $$$$\mathrm{call}\:\mathrm{l}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\mathrm{x}\right] \\ $$$$\Rightarrow\forall\varepsilon>\mathrm{0}\:\exists\eta>\mathrm{0}\mid\:\:\:\mid\mathrm{x}\mid<\eta\Rightarrow\mid\left[\mathrm{x}\right]−\mathrm{l}\mid<\varepsilon \\ $$$$\mathrm{let}\:\epsilon=\frac{\mathrm{l}}{\mathrm{3}} \\ $$$$\left.\Rightarrow\exists\eta_{\varepsilon} \:\mathrm{such}\:\mathrm{that}\:\forall\mathrm{x}\in\right]−\eta,\eta\left[\:\:\mid\left[\mathrm{x}\right]−\mathrm{L}\mid<\frac{\mathrm{1}}{\mathrm{3}}\right. \\ $$$$\left.\mathrm{for}\:\mathrm{x}\in\right]−\eta,\mathrm{0}\left[\:\Rightarrow\mid−\mathrm{1}−\mathrm{L}\mid<\frac{\mathrm{1}}{\mathrm{2}}\Leftrightarrow\mid\mathrm{1}+\mathrm{L}\mid<\frac{\mathrm{1}}{\mathrm{3}}\right. \\ $$$$\left.\mathrm{for}\:\mathrm{x}\in\right]\mathrm{0},\eta\left[\Rightarrow\mid−\mathrm{L}\mid<\frac{\mathrm{1}}{\mathrm{3}}\right. \\ $$$$\mathrm{we}\:\mathrm{have}\:\mid\mathrm{1}+\mathrm{L}−\mathrm{L}\mid=\mathrm{1} \\ $$$$\mathrm{truangular}\:\mathrm{inquality}\Rightarrow\mid\mathrm{1}\mid=\mathrm{1}\leqslant\mid−\mathrm{L}\mid+\mid\mathrm{1}+\mathrm{L}\mid\leqslant\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\mathrm{absurd}\:\mathrm{since}\:\mathrm{that}\:\mathrm{L}/\mathrm{dosent}\:\mathrm{exist} \\ $$$$ \\ $$$$\left.\mathrm{2}\right)\: \\ $$$$\mathrm{y}=\mathrm{3x}^{\mathrm{2}} +\mathrm{k}\:\:\mathrm{x}\in\left[\mathrm{k},\mathrm{k}+\mathrm{1}\left[\right.\right. \\ $$

Commented by Rio Michael last updated on 03/Nov/19

thanks

$${thanks} \\ $$

Commented by mind is power last updated on 03/Nov/19

most Welcom

$$\mathrm{most}\:\mathrm{Welcom} \\ $$

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