Let $f(z) = (\frac{z}{\bar{z}})^{2}$ , be a complex valued function , we need to prove that $\lim_{z \to 0} f(z)$ does not exists.
So , to prove that its limit doesn't exists , we approach (0,0) from 2 different paths and say that , since the limits of 2 different paths are not equal , thus the limit does'not exists. (As for the existence of the limit , limit along different paths should be equal).
So in this case we approach along the $X-axis$ and then the $Y-axis$ and see that the limit value differs ,
But the solution says , taking these paths isn't sufficient to say that the limit does'not exists. Hence in the solution the line through the origin , $Y=X$ is taken to approach the origin.
Can anyone explain me this ?
Best Answer
It's not sufficient because if you approach $0$ along the $y$-axis, the function is constant and is again $1$.
Indeed $$ \lim_{t\to0}f(it)=\lim_{t\to0}\left(\frac{it}{-it}\right)^{\!2}=1 $$
The hint is to try $$ \lim_{t\to0}f(t+it)= \lim_{t\to0}\left(\frac{t+it}{t-it}\right)^{\!2}= \left(\frac{1+i}{1-i}\right)^{\!2}=\dots $$
A different way to do it is to consider $\def\theta{\vartheta}z=re^{i\theta}$ with $\theta$ fixed and $r\to0$: then $$ \lim_{r\to0}f(re^{i\theta})= \lim_{r\to0}\left(\frac{re^{i\theta}}{re^{-i\theta}}\right)^{\!2}= e^{4i\theta} $$ which clearly depends on $\theta$.