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Question Number 189870 by uchihayahia last updated on 23/Mar/23

how do i prove this using ε and δ please help lim_(x,y→(2,1)) x^2 y=4

$$ \\ $$$$\:{how}\:{do}\:{i}\:{prove}\:{this}\:{using}\:\epsilon\:{and}\:\delta \\ $$$$\:{please}\:{help} \\ $$$$\underset{{x},{y}\rightarrow\left(\mathrm{2},\mathrm{1}\right)} {\mathrm{lim}}\:{x}^{\mathrm{2}} {y}=\mathrm{4} \\ $$$$ \\ $$

Answered by mehdee42 last updated on 23/Mar/23

we have to show : ∀ε>0 ∃ δ>0 ; ∥(x,y)−(2,1)∥<δ⇒∣x^2 y−4∣<ε if δ_1 =1 ⇒ 2<x<3 , 1<y<2 ∣x^2 y−4∣=∣xy(x−2)+2x(y−1)+2(x−2)∣−≤∣xy∣∣x−2∣+2∣x∣∣y−1∣+2∣x−2∣≤8∣x−2∣+6∣y−1∣≤ 8(√((x−2)^2 +(y−1)^2 ))+6(√((x−2)^2 +(y−1)^2 ))=14∥(x,y)−(2,1)∥<ε⇒δ_2 =(1/(14))ε ⇒δ=min{δ_1 ,δ_2 }

$${we}\:{have}\:{to}\:{show}\:: \\ $$$$\forall\epsilon>\mathrm{0}\:\exists\:\delta>\mathrm{0}\:\:;\:\parallel\left({x},{y}\right)−\left(\mathrm{2},\mathrm{1}\right)\parallel<\delta\Rightarrow\mid{x}^{\mathrm{2}} {y}−\mathrm{4}\mid<\epsilon \\ $$$${if}\:\delta_{\mathrm{1}} =\mathrm{1}\:\Rightarrow\:\mathrm{2}<{x}<\mathrm{3}\:,\:\mathrm{1}<{y}<\mathrm{2} \\ $$$$\mid{x}^{\mathrm{2}} {y}−\mathrm{4}\mid=\mid{xy}\left({x}−\mathrm{2}\right)+\mathrm{2}{x}\left({y}−\mathrm{1}\right)+\mathrm{2}\left({x}−\mathrm{2}\right)\mid−\leqslant\mid{xy}\mid\mid{x}−\mathrm{2}\mid+\mathrm{2}\mid{x}\mid\mid{y}−\mathrm{1}\mid+\mathrm{2}\mid{x}−\mathrm{2}\mid\leqslant\mathrm{8}\mid{x}−\mathrm{2}\mid+\mathrm{6}\mid{y}−\mathrm{1}\mid\leqslant \\ $$$$\mathrm{8}\sqrt{\left({x}−\mathrm{2}\right)^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} }+\mathrm{6}\sqrt{\left({x}−\mathrm{2}\right)^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{14}\parallel\left({x},{y}\right)−\left(\mathrm{2},\mathrm{1}\right)\parallel<\epsilon\Rightarrow\delta_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{14}}\epsilon \\ $$$$\Rightarrow\delta={min}\left\{\delta_{\mathrm{1}} ,\delta_{\mathrm{2}} \right\} \\ $$$$ \\ $$

Commented by uchihayahia last updated on 23/Mar/23

thank you, why ∣xy∣ became 8?

$${thank}\:{you},\:{why}\:\mid{xy}\mid\:{became}\:\mathrm{8}? \\ $$

Commented by mehdee42 last updated on 23/Mar/23

∣x∣<3&∣y∣<2⇒∣x∣∣y∣∣x−2∣<6∣x−2∣ ⇒∣x∣y∣∣x−2∣+...+2∣x−2∣<6∣x−2∣+...+2∣x−2∣=8∣x−2∣+...

$$\mid{x}\mid<\mathrm{3\&}\mid{y}\mid<\mathrm{2}\Rightarrow\mid{x}\mid\mid{y}\mid\mid{x}−\mathrm{2}\mid<\mathrm{6}\mid{x}−\mathrm{2}\mid \\ $$$$\Rightarrow\mid{x}\mid{y}\mid\mid{x}−\mathrm{2}\mid+...+\mathrm{2}\mid{x}−\mathrm{2}\mid<\mathrm{6}\mid{x}−\mathrm{2}\mid+...+\mathrm{2}\mid{x}−\mathrm{2}\mid=\mathrm{8}\mid{x}−\mathrm{2}\mid+... \\ $$

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