Deduce that if |x -3| < 1, then |(x + 2)^(2) - 25| < 11|x - 3|

Use this result to show, without using the Limit Laws, that

lim x -> 3 of (x + 2)^(2) = 25

I am really confused and dont understand this question

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- Apr 24th 2010, 01:42 AMSyNtHeSiSLimits
Deduce that if |x -3| < 1, then |(x + 2)^(2) - 25| < 11|x - 3|

Use this result to show, without using the Limit Laws, that

lim x -> 3 of (x + 2)^(2) = 25

I am really confused and dont understand this question - Apr 24th 2010, 01:52 AMsimplependulum

$\displaystyle |x-3|<1 $

$\displaystyle |(x+2)^2 - 25| = |(x+2)^2 - 5^2| = |(x+2-5)(x+2+5)|$

$\displaystyle = |(x-3)(x+7)| $

Also from $\displaystyle |x-3|<1 $ , we have $\displaystyle 2<x<4 $ so $\displaystyle 9 < x+7 < 11 $

Therefore , $\displaystyle |(x+2)^2 - 25| = |x-3||x+7| < 11|x-3|$

For the second part , we have to prove it by checking the definition ... What similarity between part 1 and the definition can you find ? - May 7th 2010, 04:06 PMSyNtHeSiS
How are you suppose to know that it is suppose to be manipulated in that way? I would never have thought of that (Surprised)