Calculate the limit. lim_((x,y)rarr(0,0))(x^(2)+y^(4))/(x^(4)+y^(2))

Check Initial Value: First, let's plug in the values of x and y as 0 to see if the function is defined at that point. lim_((x,y)rarr(0,0))(x^(2)+y^(4))/(x^(4)+y^(2)) = (0^2 + 0^4) / (0^4 + 0^2) = 0/0 We get an indeterminate form, so we need to do more work to find the limit.Approach Along y-axis: Let's try approaching (0, 0) along the y-axis (x=0). lim_((x,y)rarr(0,0))(x^(2)+y^(4))/(x^(4)+y^(2)) = lim_((y)rarr(0))(0^2+y^(4))/(0^4+y^(2)) = lim_((y)rarr(0))y^(4)/y^(2) Simplify the expression. lim_((y)rarr(0))y^(4)/y^(2) = lim_((y)rarr(0))y^(2) = 0Approach Along x-axis: Now, let's approach (0, 0) along the x-axis (y=0). lim_((x,y)rarr(0,0))(x^(2)+y^(4))/(x^(4)+y^(2)) = lim_((x)rarr(0))(x^(2)+0^(4))/(x^(4)+0^(2)) = lim_((x)rarr(0))x^(2)/x^(4) Simplify the expression. lim_((x)rarr(0))x^(2)/x^(4) = lim_((x)rarr(0))1/x^(2) As x approaches 0, 1/x^2 approaches infinity.

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Q. Calculate the limit.

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\lim _{(x, y) \rightarrow(0,0)} \frac{x^{2}+y^{4}}{x^{4}+y^{2}}

Check Initial Value: First, let's plug in the values of $x$ and $y$ as $0$ to see if the function is defined at that point. $\newline$ $\lim_{(x,y)\to(0,0)}\frac{x^{2}+y^{4}}{x^{4}+y^{2}} = \frac{0^2 + 0^4}{0^4 + 0^2} = \frac{0}{0}$ $\newline$ We get an indeterminate form, so we need to do more work to find the limit.
Approach Along y-axis: Let's try approaching $(0, 0)$ along the y-axis $(x=0)$ . $\lim_{(x,y)\to(0,0)}\frac{x^{2}+y^{4}}{x^{4}+y^{2}} = \lim_{(y)\to(0)}\frac{0^2+y^{4}}{0^4+y^{2}} = \lim_{(y)\to(0)}\frac{y^{4}}{y^{2}}$ Simplify the expression. $\lim_{(y)\to(0)}\frac{y^{4}}{y^{2}} = \lim_{(y)\to(0)}y^{2} = 0$
Approach Along x-axis: Now, let's approach $(0, 0)$ along the x-axis $(y=0)$ . $\lim_{(x,y)\to(0,0)}\frac{x^{2}+y^{4}}{x^{4}+y^{2}} = \lim_{(x)\to(0)}\frac{x^{2}+0^{4}}{x^{4}+0^{2}} = \lim_{(x)\to(0)}\frac{x^{2}}{x^{4}}$ Simplify the expression. $\lim_{(x)\to(0)}\frac{x^{2}}{x^{4}} = \lim_{(x)\to(0)}\frac{1}{x^{2}}$ As $x$ approaches $0$ , $\frac{1}{x^2}$ approaches infinity.