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Question Number 182649 by amin96 last updated on 12/Dec/22

$$\mathit{f}(\mathit{x};\mathit{y})=\frac{{\mathit{x}}^2+{\mathit{y}}^2}{{\mathit{x}}^2+{\mathit{y}}^4}$$$$\underset{x\to \mathrm{\infty }}{\lim }\left(\underset{y\to 0}{\lim }\mathit{f}(\mathit{x};\mathit{y})\right)=?$$$$\underset{\mathit{y}\to \mathrm{\infty }}{\lim }\left(\underset{x\to \mathrm{\infty }}{\lim }\mathit{f}(\mathit{x};\mathit{y})\right)=?$$

Answered by FelipeLz last updated on 12/Dec/22

$$\underset{x\to \mathrm{\infty }}{\lim }\left(\underset{y\to 0}{\lim }\frac{x^2+y^2}{x^2+y^4}\right)=\underset{x\to \mathrm{\infty }}{\lim }\frac{x^2}{x^2}=\underset{x\to \mathrm{\infty }}{\lim 1}=1$$$$\underset{y\to \mathrm{\infty }}{\lim }\left(\underset{x\to \mathrm{\infty }}{\lim }\frac{x^2+y^2}{x^2+y^4}\right)=\underset{y\to \mathrm{\infty }}{\lim }\left(\underset{x\to \mathrm{\infty }}{\lim }\frac{(x^2+y^2)x^{-2}}{(x^2+y^4)x^{-2}}\right)=\underset{y\to \mathrm{\infty }}{\lim }\left(\underset{x\to \mathrm{\infty }}{\lim }\frac{1+y^2{x}^{-2}}{1+y^4{x}^{-2}}\right)=\underset{y\to \mathrm{\infty }}{\lim }\left(\frac{1}{1}\right)=1$$

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