\( \lim _ { ( x , y ) \rightarrow ( 0,0 ) } \frac { 2 x y ^ { 2 } } { 3 x ^ { 2 } + y ^ { 4 } } \)

To evaluate the limit \( \lim _ { ( x , y ) \rightarrow ( 0,0 ) } \frac { 2 x y ^ { 2 } } { 3 x ^ { 2 } + y ^ { 4 } } \), we can investigate the behavior of the function along various paths approaching \((0, 0)\). 1. **Approaching along the line \( y = kx \)** (where \( k \) is a constant): \[ \frac{2x(kx)^2}{3x^2 + (kx)^4} = \frac{2kx^3}{3x^2 + k^4x^4} = \frac{2k x^3}{x^2(3 + k^4 x^2)} = \frac{2k x}{3 + k^4 x^2} \] As \( x \rightarrow 0 \), this approaches \( 0 \). 2. **Approaching along the line \( x = 0 \)**: \[ \frac{2(0)y^2}{3(0)^2 + y^4} = \frac{0}{y^4} = 0 \] 3. **Approaching along the line \( y = 0 \)**: \[ \frac{2x(0)^2}{3x^2 + (0)^4} = \frac{0}{3x^2} = 0 \] 4. **Approaching along the path \( y = x^n \) for \( n > 0 \)**: \[ \frac{2x(x^n)^2}{3x^2 + (x^n)^4} = \frac{2x^{2+n}}{3x^2 + x^{4n}} \] Factoring out \( x^2 \) gives: \[ = \frac{2x^{2+n}}{x^2(3 + x^{4n-2})} = \frac{2x^n}{3 + x^{4n-2}} \] For \( n = 1 \), as \( x \rightarrow 0 \) this approaches \( 0 \). For other \( n \) values, we similarly see the limit will approach \( 0 \). Since all paths approached yielded \( 0 \), we conclude that: \[ \lim _ { ( x , y ) \rightarrow ( 0,0 ) } \frac { 2 x y ^ { 2 } } { 3 x ^ { 2 } + y ^ { 4 } } = 0. \]