What is the limit as x,y approaches (0,0) of ((2x^2y))/(x^2+y^4) ?

The limit as x,y approaches (0,0) of ((2x^2y))/(x^2+y^4) is 0

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limit as x,y approaches (0,0) of ((2x^2y))/(x^2+y^4)

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

0

Solution steps

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

Convert to polar coordinates

x = r cos (θ)

= r \to 0 lim (\frac{2 ( r cos ( θ ) ) ^{2} r sin ( θ )}{( r cos ( θ ) ) ^{2} + ( r sin ( θ ) ) ^{4}})

Simplify

\frac{2 ( r cos ( θ ) ) ^{2} r sin ( θ )}{( r cos ( θ ) ) ^{2} + ( r sin ( θ ) ) ^{4}} : \frac{2 r cos ^{2} ( θ ) sin ( θ )}{r ^{2} sin ^{4} ( θ ) + cos ^{2} ( θ )}

= r \to 0 lim (\frac{2 r cos ^{2} ( θ ) sin ( θ )}{r ^{2} sin ^{4} ( θ ) + cos ^{2} ( θ )})

Plug in the value

r = 0

= \frac{2 \cdot 0 \cdot cos ^{2} ( θ ) sin ( θ )}{0 ^{2} sin ^{4} ( θ ) + cos ^{2} ( θ )}

Simplify

\frac{2 \cdot 0 \cdot cos ^{2} ( θ ) sin ( θ )}{0 ^{2} sin ^{4} ( θ ) + cos ^{2} ( θ )} : 0

= 0

x \to 2 + lim (3 - x)

\frac{d}{d x} (\frac{2 ^{x}}{cos ( x )})

y^{^{''}} + 2 y^{^{'}} + y = 64 e^{(7 t)}

\frac{d}{d x} ((x - 3)^{- 2})

\frac{\partial}{\partial x} (\frac{2 x}{2 y})

What is the limit as x,y approaches (0,0) of ((2x^2y))/(x^2+y^4) ?
The limit as x,y approaches (0,0) of ((2x^2y))/(x^2+y^4) is 0