Consider
lim ( x , y ) → ( 0 , 0 ) 6 x 2 y 3 x 2 + y 3 \lim\limits_{(x,y)\to(0,0)}{6x^2y^3 \over x^2+y^3} ( x , y ) → ( 0 , 0 ) lim x 2 + y 3 6 x 2 y 3 Suppose ε > 0 \varepsilon>0 ε > 0
∣ 6 x 2 y 3 x 2 + y 3 ∣ = 6 ∣ y 3 x 2 + y 3 ∣ x 2 \bigg|{6x^2y^3 \over x^2+y^3}\bigg|=6\bigg|{y^3 \over x^2+y^3}\bigg|x^2 ∣ ∣ x 2 + y 3 6 x 2 y 3 ∣ ∣ = 6 ∣ ∣ x 2 + y 3 y 3 ∣ ∣ x 2 Note that
∣ y 3 x 2 + y 3 ∣ ≤ 1 , x 2 ≤ x 2 + y 2 = δ 2 \bigg|{y^3 \over x^2+y^3}\bigg|\leq1, x^2\leq x^2+y^2=\delta^2 ∣ ∣ x 2 + y 3 y 3 ∣ ∣ ≤ 1 , x 2 ≤ x 2 + y 2 = δ 2 So
6 ∣ y 3 x 2 + y 3 ∣ x 2 < 6 ⋅ 1 ⋅ δ 2 6\bigg|{y^3 \over x^2+y^3}\bigg|x^2<6\cdot1\cdot\delta^2 6 ∣ ∣ x 2 + y 3 y 3 ∣ ∣ x 2 < 6 ⋅ 1 ⋅ δ 2 If we choose δ = ε 6 \delta=\sqrt{\dfrac{\varepsilon}{6}} δ = 6 ε then
6 ∣ y 3 x 2 + y 3 ∣ x 2 < 6 ⋅ 1 ⋅ δ 2 = ε 6\bigg|{y^3 \over x^2+y^3}\bigg|x^2<6\cdot1\cdot\delta^2=\varepsilon 6 ∣ ∣ x 2 + y 3 y 3 ∣ ∣ x 2 < 6 ⋅ 1 ⋅ δ 2 = ε
∣ f ( x , y ) − 0 ∣ < ε , i f x 2 + y 2 < δ |f(x,y)-0|<\varepsilon, if\ \sqrt{x^2+y^2}<\delta ∣ f ( x , y ) − 0∣ < ε , i f x 2 + y 2 < δ Hence
lim ( x , y ) → ( 0 , 0 ) 6 x 2 y 3 x 2 + y 3 = 0 \lim\limits_{(x,y)\to(0,0)}{6x^2y^3 \over x^2+y^3}=0 ( x , y ) → ( 0 , 0 ) lim x 2 + y 3 6 x 2 y 3 = 0
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