Find limlimits_(z→1)(√(z^2+3)-2)/(z-1) , where √(z^2+3) is…

Find $\ \lim\limits_{z\to1}\dfrac{\sqrt{z^2+3}-2}{z-1}$, where $\sqrt{z^2+3}$ is the continuous branch with $f(0)=\sqrt3$.

Step 1 · Confirm the indeterminate form

At $z=1$: $\sqrt{1+3}-2=2-2=0$, and $1-1=0$.

Step 2 · Rationalize

$$ \frac{\sqrt{z^2+3}-2}{z-1}\cdot\frac{\sqrt{z^2+3}+2}{\sqrt{z^2+3}+2}=\frac{(z^2+3)-4}{(z-1)(\sqrt{z^2+3}+2)}=\frac{z^2-1}{(z-1)(\sqrt{z^2+3}+2)} $$

Step 3 · Factor and cancel

$$ =\frac{(z-1)(z+1)}{(z-1)(\sqrt{z^2+3}+2)}=\frac{z+1}{\sqrt{z^2+3}+2} \qquad(z\neq1) $$

Step 4 · Evaluate the limit

$$ \boxed{\lim_{z\to1}\frac{z+1}{\sqrt{z^2+3}+2}=\frac{2}{2+2}=\frac12} $$

Branch check

The condition $f(0)=\sqrt3>0$ selects the branch that stays positive on the real axis near $0$; since $z^2+3$ never vanishes for real $z$, this branch stays continuous and positive all the way to $z=1$, giving $\sqrt{1^2+3}=+2$ as used above.

Cross-check · derivative interpretation

This limit is exactly $f'(1)$ for $f(z)=\sqrt{z^2+3}$: by the chain rule $f'(z)=\dfrac z{\sqrt{z^2+3}}$, so $f'(1)=\dfrac1{\sqrt4}=\dfrac12$ — matching Step 4 directly.