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I think this is a limit proof question, but I wasn't sure. Any help?

Use a graph to find the largest number $\delta > 0$ such that if $|x-1|<\delta$, then $|x^3-5x+6-2| < \varepsilon$ when $\varepsilon=0.2$

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Yes, this is a limit proof question. – apnorton Jan 15 '13 at 20:48
Did you plot $|x^3-5x+6-2| < \varepsilon $ yet? – rschwieb Jan 15 '13 at 20:49
The question is about limits, but there is no sign of a proof... – rschwieb Jan 15 '13 at 20:50
up vote 2 down vote accepted

Here's the graph of $y = x^3-5x+6$ as well as $y=2$ and $y=2\pm\varepsilon$. Can you read off the answer?

enter image description here

If you find the first graph difficult to read, here's $y = |x^3-5x+6-2|$ together with $y=\varepsilon$.

enter image description here

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ooohh thank you! so the answer would be anything between .911<delta<1.124? – user57172 Jan 15 '13 at 20:59
@user57172 No, $x$ can be between $0.911$ and $1.124$. What does that say about $\delta$? – mrf Jan 15 '13 at 21:02
would I have to subtract 1 from those? – user57172 Jan 15 '13 at 21:03
@user57172 $-0.089 < x-1 < 0.124$, so how large can you choose $\delta$ to ensure that $-\delta < x-1 < \delta$? – mrf Jan 15 '13 at 21:06
you can choose up to .124? – user57172 Jan 15 '13 at 21:10

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