I need help with a limit proof

Question

I tried to proof this limit but when i get epsilon i can't narrow it because there's a factorization something unusual. The limit that i need to proof is: $\lim_{z\to1}\frac{{z^2 -1}}{z-1} = 2$ I would appreciate if somebody can help me with this proof. Thank you.

Answer 1

For every $z\ne1$, $\displaystyle\left|\frac{z^2-1}{z-1}-2\right|=\left|z-1\right|$. If $z\to1$, the RHS converges to zero. QED.

Answer 2

$$\large z^2-1=(z-1)(z+1)$$ $$f(z)=\frac{z^2-1}{z-1}=\begin{cases}\begin{align}z+1\quad z\ne1\\\text{undefined}\quad z=1\end{align}\end{cases}$$ For every $\epsilon>0$, there is some number $\delta>0$ such that: $$|f(z)-2|<\epsilon\qquad\text{whenever}\qquad 0<|z-1|<\delta$$ $\left(\text{actually }\delta(\epsilon)=\epsilon\right)$

which is equivalent saying: $$\lim_{z\to1}f(z)=2$$