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Does the following limit exist? $$\lim_{z\rightarrow 0}\frac{Re(z)^2}{|z|^2}$$ Here, Re means the real part of the function.

This is what I have so far:

In order for the limit to exist, z must be allowed to approach $0$ from any direction.

With z approaching $0$ through values $z=x+i0$: $$\lim_{x\rightarrow 0}\frac{x^2-0}{x^2-0}=1$$ With z approaching $0$ through values $z=iy$ : $$\lim_{y\rightarrow 0}\frac{-y^2-0}{y^2-0}=-1$$ Since the two limits are different, the limit does not exist.

Does this logic make sense?

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Let $x=\text{Re}(z)$ and $y=\text{Im}(z)$, then we have

$$\frac{\left(\text{Re}(z)\right)^2}{|z|^2}=\frac{x^2}{x^2+y^2}$$

while

$$\frac{\left(\text{Re}(z^2)\right)}{|z|^2}=\frac{x^2-y^2}{x^2+y^2}$$

Neither $\lim_{(x,y)\to(0,0)}\frac{x^2}{x^2+y^2}$ nor $\lim_{(x,y)\to(0,0)}\frac{x^2-y^2}{x^2+y^2}$ exists.

One can show that these limits fail to exist by analyzing the behaviors along $x=0$ and $y=0$ separately.

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The idea of approaching $0$ along different lines is good, but your second calculation contains a mistake. If $z=iy$ (I'm assuming you mean $y$ real here, then $\operatorname{Re}(z)=0$.

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Yes the idea is correct, that the standard way to show that a limit does not exist, indeed we have that

  • for x=y $\frac{\left(\text{Re}(z)\right)^2}{|z|^2}=\frac{x^2}{x^2+x^2}\to \frac12$

  • for x=o $\frac{\left(\text{Re}(z)\right)^2}{|z|^2}=\frac{o}{o+y^2}\to 0$

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