Prove that the limit $\displaystyle\lim_{z\to 0} \frac{|z|^{2a}+i(\operatorname{Re}(z))^a}{z},\ 0<a<1$ does not exist.

Question

I am trying to prove that the limit $$\lim_{z\to 0} \frac{|z|^{2a}+i(\operatorname{Re}(z))^a}{z},\ 0<a<1$$ does not exist.

My argument: Let $z=x>0$ (real number positive number), then the previous limit is equal to $\displaystyle\lim_{x\to 0} \ (x^{2a-1}+ix^{a-1})$ and consequently $\lim_{x\to 0} x^{a-1}$ is not a real number. So, the limit does not exist for all $a\in (0,1)$. Is it right?? Thanks.

Answer 1

That is a good approach. In order to eliminate that absolute value, it is better to restrict $x$ to the non-negative real numbers. There are two possibilites then:

• $0<a<\frac12$: then $ix^{a-1}$ takes purely imaginary values with arbitrarily large absolute value and $x^{2a-1}$ takes real values with arbitrarily large absolute value. So, the limit $\lim_{x\to0^+}x^{2a-1}+ix^{a-1}$ does not exist.
• $\frac12\leqslant a<1$: then $ix^{a-1}$ takes purely imaginary values with arbitrarily large absolute value and $\lim_{x\to0^+}x^{2a-1}\in\Bbb R$. So, again, the limit $\lim_{x\to0^+}x^{2a-1}+ix^{a-1}$ does not exist.

Answer 2

Here is another way to approach the problem. With our loan of generality, assumption $r=|z|<1$. If one uses polar coordinates, the expression in your limit becomes

$$\frac{r^{2a} +ir^a|\cos\theta|^a}{re^{i\theta}}=r^{a-1}e^{-i\theta}\Big(r^a+i|\cos\theta|^a\Big)$$

There factor $r^{a-1}\xrightarrow{r\rightarrow0}\infty$, while the factor $(r^a+ I|\cos\theta|^a)$ is bounded away from $0$ for say $[0,\pi/2-\varepsilon]\cup[\pi/2+\varepsilon,3\pi/2-\varepsilon]\cup[3\pi/2+\varepsilon,2\pi]$ for any fixed small $\varepsilon>0$. If draw a picture, the region I'm describing is the interior of the circle where the inside of the circular sectors within an $\epsilon$ angle with respect to the $y$-axis.