Does the function f(z)=(1-|z|)/(1-z) have a limit at the…

Does the function $\ f(z)=\dfrac{1-|z|}{1-z}\ $ have a limit at the point $z=1$?

Step 1 · Approach along the real axis

For real $x$ near $1$ (so $x>0$, $|x|=x$): $$ f(x)=\frac{1-x}{1-x}=1\qquad\text{whether }x\to1^-\text{ or }x\to1^+ $$

Step 2 · General ray $z=1+re^{i\theta}$, $r\to0^+$

$$ |z|^2=1+2r\cos\theta+r^2\ \Longrightarrow\ |z|\approx1+r\cos\theta\ \Longrightarrow\ 1-|z|\approx-r\cos\theta $$

while $1-z=-re^{i\theta}$ exactly, so $$ f(z)\approx\frac{-r\cos\theta}{-re^{i\theta}}=\cos\theta\, e^{-i\theta} $$

Step 3 · Evaluate along specific directions

$$ \theta=0:\ \ \cos0\cdot e^0=1;\qquad \theta=\frac\pi2:\ \ \cos\frac\pi2\cdot e^{-i\pi/2}=0;\qquad \theta=\frac\pi4:\ \ \frac{\sqrt2}2\,e^{-i\pi/4}=\frac12-\frac i2 $$

Verified numerically at $r=10^{-6}$: the ray $\theta=\pi/2$ gives $f\approx-5\times10^{-7}i\to0$, the ray $\theta=\pi/4$ gives $f\approx0.500-0.500i$, and $\theta=\pi/3$ gives $f\approx0.250-0.433i$, matching $\cos\theta\,e^{-i\theta}=\tfrac14-\tfrac{\sqrt3}4i$ — each ray converges cleanly, but to a different value.

Conclusion

Since the directional limit $\cos\theta\,e^{-i\theta}$ genuinely varies with $\theta$ (equal to $1$ along the real axis, $0$ along the imaginary direction, and everything between), the two-dimensional limit fails to exist.

Why this happens

The numerator involves $|z|$, which is real-valued and not complex-differentiable; mixing it with the holomorphic factor $1-z$ in the denominator produces genuine angular dependence that a purely holomorphic quotient could never exhibit.