Talk:Infinite Limit Theorem

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Surely you also have to specify that $h(x) \ne 0$ when $x \ne c$? --prime mover 14:33, 22 November 2011 (CST)

You're right, corrected --GFauxPas 15:03, 22 November 2011 (CST)

$\forall x \in \mathbb I: g \left({x}\right) \ne 0$ may be unnecessary, it might be sufficient that $g(c) \ne 0$, I have to read up on $\epsilon - M$ proofs and try to figure this out --GFauxPas 17:25, 22 November 2011 (CST)

Question

Why do we need $\epsilon_1 > \epsilon_2$? --GFauxPas (talk) 16:57, 20 November 2012 (UTC)

So that $\displaystyle \forall x: \left\vert{x - c}\right\vert < \epsilon_2 \implies \left\vert{x - c}\right\vert < \epsilon_1 \implies \left\vert{g \left({x}\right)}\right\vert > \frac {\left\vert{g \left({c}\right)}\right\vert} 2$. --abcxyz (talk) 17:07, 20 November 2012 (UTC)