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Mathematics 22 Online
OpenStudy (anonymous):

Use the formal definition of limit to verify the indicated limit.

OpenStudy (anonymous):

I understand the steps up to the red line.

OpenStudy (anonymous):

0<x-2<d |x+5||x-2|<E this is as far as I can get.

OpenStudy (anonymous):

|x+5|<8 I think

ganeshie8 (ganeshie8):

thats because we are assuming |x-2| < 1 so x < 3

OpenStudy (anonymous):

ok

ganeshie8 (ganeshie8):

\(\large |x^2+3x-10|=|x+5|\cdot |x-2| \lt 8 \cdot \delta \) yes ?

OpenStudy (anonymous):

yeah because |x+5|<8

ganeshie8 (ganeshie8):

yes and in the start, we have \(\delta = \min(\epsilon/8, ~1)\)

OpenStudy (anonymous):

yeah, but why does he pic that?

ganeshie8 (ganeshie8):

since putting \(\large \delta =1\) makes the value of \(\epsilon/8 \) greater than 1, the value of \(\large \delta \) must be \(\large \epsilon/8\) here

ganeshie8 (ganeshie8):

factoring |f(x) - L| must have given him that hint

OpenStudy (anonymous):

are you able to say: \[|x+5||x-2|<\epsilon \] \[|x+5|<8\] \[|x-2|<\epsilon /8\]

ganeshie8 (ganeshie8):

that looks good to me!

OpenStudy (anonymous):

Ok :) thnx again

ganeshie8 (ganeshie8):

np :)

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