91Ó°ÊÓ

To find out if the function is continuous at \((0,0)\), we need to evaluate the following limit: $$\lim_{(x,y) \to (0,0)} \frac{y^4 - 2x^2}{y^4 + x^2}.$$ To check the limit at \((0,0)\), we can approach the origin from different directions. Let \(y = mx\), where \(m\) is a real constant, and calculate the limit along this line. $$\lim_{x \to 0} \frac{(mx)^4 - 2x^2}{(mx)^4 + x^2} = \lim_{x \to 0} \frac{m^4x^4 - 2x^2}{m^4x^4 + x^2} = \lim_{x \to 0} \frac{x^2(m^4x^2 - 2)}{x^2(m^4x^2 + 1)} = \lim_{x \to 0} \frac{m^4x^2 - 2}{m^4x^2 + 1}.$$ As \(x \to 0\), both the numerator and the denominator of the fraction tend to 0. The limit will therefore depend on the value of \(m\). Since the limit depends on the direction we approach \((0,0)\), the function is not continuous at \((0,0)\).

For the rest of the points \((x, y) \neq (0,0)\), the function is given as: $$f(x, y) = \frac{y^4 - 2x^2}{y^4 + x^2}.$$ To prove continuity of the function at these points, we will use the definition of continuity which states that a function \(f(x, y)\) is continuous at a point \((a, b)\) if: $$\lim_{(x,y) \to (a,b)} f(x, y) = f(a, b).$$ Let \((a, b) \neq (0,0)\), then we have: $$\lim_{(x,y) \to (a,b)} \frac{y^4 - 2x^2}{y^4 + x^2} = \frac{b^4 - 2a^2}{b^4 + a^2} = f(a, b).$$ Since the limit exists and is equal to the function value at any point \((a, b) \neq (0,0)\), the function is continuous for all \((x, y) \neq (0,0)\). In conclusion, the given function is continuous for all points \((x, y) \in \mathbb{R}^2\) except at \((0,0)\).