<s>[INST] Limit in complex space How to prove that the limit in the complex space:
$$\lim\limits_{z \to 0} \frac{|1+z|-1}{z}=1$$
Please help me
 [/INST] $$\lim\limits_{z \to 0} \frac{|1+z|-1}{z}=\lim\limits_{(x,y) \to (0,0)} \frac{\sqrt{(x+1)^2+y^2}-1}{x+iy}$$ 
if $x=y$
$$\lim\limits_{z \to 0} \frac{|1+z|-1}{z}=\lim\limits_{x\to 0} \frac{\sqrt{(x+1)^2+x^2}-1}{x+ix}\times \frac{\sqrt{(x+1)^2+x^2}+1}{\sqrt{(x+1)^2+x^2}+1}=\lim\limits_{x\to 0} \frac{{(x+1)^2+x^2}-1}{{(x+ix)(\sqrt{(x+1)^2+x^2}+1)}}=\lim\limits_{x\to 0} \frac{x(2x+2)}{x(1+i)\sqrt{(x+1)^2+x^2}+1}=\frac{2}{(1+i)\sqrt{(0+1)^2+0^2}+1}=\frac{1}{(1+i)} $$ 
 similarly for $y=-x$
$$\lim\limits_{(x,y) \to (0,0)} \frac{\sqrt{(x+1)^2+y^2}-1}{x+iy} =\frac{1}{(1-i)}$$ 
hence the limit doesn't exist.
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