how-do-i-prove-this-using-and-please-help-lim-x-y-2-1-x-2-y-4-

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

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 ∣ + \dots + 2 ∣ x - 2 ∣< 6 ∣ x - 2 ∣ + \dots + 2 ∣ x - 2 ∣= 8 ∣ x - 2 ∣ + \dots