Prove that the function f(z) defined by $$f(z)=\frac{x^3(1+i)-y^3(1-i)}{x^2 + y^2} ; z\neq{0};f(0)=0$$
is continuous and the Cauchy Riemann equations are satisfied at the origin $(0,0)$, yet $f'(0)$ does not exist.
I am undone as to conclusively proving that the limit at origin equates 0 (which in-turn will prove the continuity). Taking $y$ as a linear/parabolic function of $x $ and moving along that path leads to the same variable-independent limit but there exists independent paths and we can't (ever) disprove that we have starved all paths:-(
Applying delta-epsilon definition is a way-out but seems to be way overboard (~8 marks is allocated and a time of around 12 minutes.)
The need of determining the particular limit will also play a final role in determining the derivative of$f(z)$ which does come out to be path-dependent; shall the above limit be $0$.
We have $|f(z)| \le \frac{\sqrt{2}(|x|^3+|y|^3)}{x^2+y^2}$ for $z \ne 0.$
Now show that $|f(z)| \le \sqrt{2}(|x|+|y|)$.
This shows that $f(z) \to 0$ as $z \to 0.$