-1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Start by factorising $z^2-1$ $\endgroup$ – almagest Sep 6 '14 at 12:49
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ What's wrong with simply simplifying $\frac{z^2-1}{z-1}$ to $z+1$? $\endgroup$ – barak manos Sep 6 '14 at 12:51
  • $\begingroup$ @barakmanos Where did I say it was? $\endgroup$ – Did Sep 6 '14 at 12:51
  • 2
    $\begingroup$ 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 $1$. $\endgroup$ – Did Sep 6 '14 at 13:01
  • 1
    $\begingroup$ @GitGud Ah OK, thanks. Then, no, the minimum is not necessary and the choice $\delta=\varepsilon$ is all right. $\endgroup$ – Did Sep 6 '14 at 13:09
  • 1
    $\begingroup$ @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. $\endgroup$ – Git Gud Sep 6 '14 at 13:13
1
$\begingroup$

$$\large z^2-1=(z-1)(z+1)$$ $$f(z)=\frac{z^2-1}{z-1}=\begin{cases}\begin{align}z+1\quad z\ne1\\\text{undefined}\quad z=1\end{align}\end{cases}$$ 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: $$\lim_{z\to1}f(z)=2$$

$\endgroup$
  • 1
    $\begingroup$ Sorry but to consider some $\varepsilon(\delta)$ reflects a deep misconception about the notion of limit. $\endgroup$ – Did Sep 6 '14 at 13:04
  • $\begingroup$ @Did what's wrong with it, seems correct to me? $\endgroup$ – RE60K Sep 6 '14 at 13:10
  • 1
    $\begingroup$ $\delta$ depends on $\epsilon$, not vice versa. $\endgroup$ – paw88789 Sep 6 '14 at 13:11
  • $\begingroup$ @Did Oh!, a typo $\endgroup$ – RE60K Sep 6 '14 at 13:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.