How do you prove that the limit of $(1-5x) = 16$ as x approaches -3 using the epsilon delta proof?

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Answer 1 · 2016-10-01T02:22:49

See proof below.

Given a positive $epsilon$ , let $delta = epsilon/5$ .

Now for every $x$ with $0 < abs(x-(-3)) < delta$ , we have

$0 < abs(x+3) < delta$ and

$abs((1-5x)-16) = abs(-5-15)$

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

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

$= 5abs(x+3)$

$< 5delta = 5(epsilon/5) = epsilon$

We have shown that for any positive $epsilon$ , there is a positive $delta$ such that for every $x$ ,

if $0 < abs(x-(-3)) < delta$ , then $abs((1-5x)-16) < epsilon$ .

Therefore, by the definition of limit, $lim_(xrarr-3)(1-5x) = 16$ .

How do you prove that the limit of (1-5x) = 16(1-5x) = 16 as x approaches -3 using the epsilon delta proof?