How do you prove that the limit of $5 - 2x$ as x approaches 2 is equal to 1 using the epsilon delta proof?

Question

2984 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-14T17:43:26

Given a positive $epsilon$ , let $delta = epsilon/2$ . Note that $delta$ is positive.

Now for every $x$ with $0 < abs(x-2) < delta$ , we have

$abs((5-2x)-1) = abs(4-2x)$

$= abs((-2)(x-2))$

$= abs(-2)abs(x-2)$

$= 2abs(x-2)$

$< 2delta$

$= 2(epsilon/2)$

$= epsilon$ .

That is: for every $x$ with $0 < abs(x-2) < delta$ , we have $abs((5-2x)-1) < epsilon$ .

So by the definition of limit, $lim_(xrarr2)(5-2x) = 1$ .

How do you prove that the limit of 5 - 2x5 - 2x as x approaches 2 is equal to 1 using the epsilon delta proof?