# I need help with a limit proof

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.

• Start by factorising $z^2-1$ – almagest Sep 6 '14 at 12:49

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.

• What's wrong with simply simplifying $\frac{z^2-1}{z-1}$ to $z+1$? – barak manos Sep 6 '14 at 12:51
• @barakmanos Where did I say it was? – Did Sep 6 '14 at 12:51
• egarro: Let $f$ denote the function of interest. Indeed, choosing $\delta=\varepsilon$ yields $$|z-1|\leqslant\delta,z\ne1\implies\left|f(z)-2\right|\leqslant\varepsilon.$$ Note that the condition $z\ne1$ is necessary because $f$ is not defined at $1$. – Did Sep 6 '14 at 13:01
• @GitGud Ah OK, thanks. Then, no, the minimum is not necessary and the choice $\delta=\varepsilon$ is all right. – Did Sep 6 '14 at 13:09
• @barakmanos Your method of proof is something that would only come up after you proved some things with $\varepsilon$-$\delta$ and not only does the OP want an $\varepsilon$-$\delta$ proof, this answer directly hints at how to get one due the equality stated in it. – Git Gud Sep 6 '14 at 13:13

$$\large z^2-1=(z-1)(z+1)$$ 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: $$\lim_{z\to1}f(z)=2$$

• Sorry but to consider some $\varepsilon(\delta)$ reflects a deep misconception about the notion of limit. – Did Sep 6 '14 at 13:04
• @Did what's wrong with it, seems correct to me? – RE60K Sep 6 '14 at 13:10
• $\delta$ depends on $\epsilon$, not vice versa. – paw88789 Sep 6 '14 at 13:11
• @Did Oh!, a typo – RE60K Sep 6 '14 at 13:12