how do i prove this using ε and δ please help lim


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

$h o w d o i p r o v e t h i s u s i n g ϵ a n d δ$ $p l e a s e h e l p$ $\lim_{x, y \to (2, 1)} x^{2} y = 4$
	Answered by mehdee42 last updated on 23/Mar/23
	$we have to show : ∀ε>0 ∃ δ>0 ; ∥(x,y)−(2,1)∥<δ⇒∣x^2 y−4∣<ε if δ_1 =1 ⇒ 2<x<3 , 1<y<2 ∣x^2 y−4∣=∣xy(x−2)+2x(y−1)+2(x−2)∣−≤∣xy∣∣x−2∣+2∣x∣∣y−1∣+2∣x−2∣≤8∣x−2∣+6∣y−1∣≤ 8(√((x−2)^2 +(y−1)^2 ))+6(√((x−2)^2 +(y−1)^2 ))=14∥(x,y)−(2,1)∥<ε⇒δ_2 =(1/(14))ε ⇒δ=min{δ_1 ,δ_2 }$
	$w e h a v e t o s h o w :$ $\forall ϵ > 0 \exists δ > 0; ∥ (x, y) - (2, 1) ∥< δ \Rightarrow∣ x^{2} y - 4 ∣< ϵ$ $i f δ_{1} = 1 \Rightarrow 2 < x < 3, 1 < y < 2$ $∣ x^{2} y - 4 ∣=∣ x y (x - 2) + 2 x (y - 1) + 2 (x - 2) ∣ - ⩽∣ x y ∣∣ x - 2 ∣ + 2 ∣ x ∣∣ y - 1 ∣ + 2 ∣ x - 2 ∣⩽ 8 ∣ x - 2 ∣ + 6 ∣ y - 1 ∣⩽$ $8 \sqrt{{(x - 2)}^{2} + {(y - 1)}^{2}} + 6 \sqrt{{(x - 2)}^{2} + {(y - 1)}^{2}} = 14 ∥ (x, y) - (2, 1) ∥< ϵ \Rightarrow δ_{2} = \frac{1}{14} ϵ$ $\Rightarrow δ = m i n {δ_{1}, δ_{2}}$
	Commented by uchihayahia last updated on 23/Mar/23

	$t h a n k y o u, w h y ∣ x y ∣ b e c a m e 8 ?$
	Commented by mehdee42 last updated on 23/Mar/23

	$∣ x ∣< 3 & ∣ y ∣< 2 \Rightarrow∣ x ∣∣ y ∣∣ x - 2 ∣< 6 ∣ x - 2 ∣$ $\Rightarrow∣ x ∣ y ∣∣ x - 2 ∣ + . . . + 2 ∣ x - 2 ∣< 6 ∣ x - 2 ∣ + . . . + 2 ∣ x - 2 ∣= 8 ∣ x - 2 ∣ + . . .$
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