How do you prove the statement lim as x approaches 3 for $(x^2+x-4) = 8$ using the epsilon and delta definition?

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Answer 1 · 2015-09-19T03:41:30

Please see the explanation section.

We want to make $abs((x^2+x-4)-8) < epsilon$

We control $delta$ and $abs(x-3) < delta$

Note that:

$abs((x^2+x-4)-8) =abs(x^2+x-12)$

$= abs((x+4)(x-3))$

$= abs(x+4) abs(x-3)$

We control $abs(x-3)$ and so, indirectly, we control $abs(x+4)$

Make $abs(x-3) < 1$ . (Any positive number will work in place of $1$ , but the details of what follows will change.)

This assures us that $2 < x < 4$ .

So considering $x+4$ ( the other factor), we see:

$6 < x+4 < 7.$ And

$abs(x+4) < 7$ .

So, if $abs(x-3) < 1$ , then

$abs((x^2+x-4)-8) = abs(x+4) abs(x-3) < 7abs(x-3)$

If we also make sure that $abs(x-3) < epsilon/7$ , then we will have:

$abs((x^2+x-4)-8) < 7abs(x-3) < 7(epsilon/7) = epsilon$ .

Now write the proof :

Given $epsilon > 0$ , let $delta =min {1, epsilon/7}$ .

Now is $abs(x-3) < delta$ , we will have:

$abs((x^2+x-4)-8) = abs(x^2+x-12) = abs(x+4) abs(x-3)$

$< 7 (epsilon/7) = epsilon$

How do you prove the statement lim as x approaches 3 for (x^2+x-4) = 8(x^2+x-4) = 8 using the epsilon and delta definition?