How do you prove the statement lim as x approaches 2 for $((x^2+x-6)/(x-2))=5$ using the epsilon and delta definition?

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Answer 1 · 2015-10-16T16:33:53

See the explanation below.

Preliminary Analysis

First look at $abs(f(x)-L)$ . I want to make that less than $epsilon$ .

I am looking for $abs(x-2)$ because, through my control of $delta$ , I control $abs(x-2)$ .

$abs(f(x)-L) = abs((x^2+x-6)/(x-2)-5)$

$= abs(((x+3)(x-2))/(x-2)-5)$

$= abs((x+3)-5)$ for all $x != 2$

$= abs(x-2)$ for all $x != 2$

We want to make $abs(f(x)-L)$ less than a given $epsilon$ .

If we make $abs(x-2) < epsilon$ BUT $x != 2$ , then we will have what we want. (We cannot allow $x=2$ because $2$ is not in the domain of $f$ .

So now, we are ready to write the proof.

Proof

Given $epsilon > 0$ choose $delta = epsilon$ . (Clearly, then, $delta > 0$ as required.

Now if $x$ is chosen so that

$0 < abs(x-2) < delta$ , then we will have

$abs((x^2+x-6)/(x-2)-5) = abs(((x+3)(x-2))/(x-2)-5)$

$= abs((x+3)-5) = abs(x-2) < delta = epsilon$ .

That is:
If $0 < abs(x-2) < delta$ , then $abs((x^2+x-6)/(x-2)-5) < epsilon$ .

By the definition of limit,

$lim_(xrarr2)(x^2+x-6)/(x-2) = 5$

How do you prove the statement lim as x approaches 2 for ((x^2+x-6)/(x-2))=5((x^2+x-6)/(x-2))=5 using the epsilon and delta definition?