How do you use the epsilon delta definition to prove that the limit of $2x-4=6$ as $x->1$ ?

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Answer 1 · 2017-05-09T13:11:26

It cannot be proven. It is false.

$lim_(xrarr1)(2x-4) = -2$ and $-2 != 6$

Proof that $lim_(xrarr1)(2x-4) = -2$

Given $epsilon > 0$ , choose $delta = epsilon/2$ (Note that $delta > 0$ .)

Then for any $x$ with $0 < abs (x-1) < delta$ ,

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

$= 2abs(x-1)$

$< 2delta$

$= 2epsilon/2$

$= epsilon$

That is: $abs(2x-4)-(-2)) < epsilon$ .

We have shown that:

For any $epsilon > 0$ , there is a $delta > 0$ such that

for all $x$ with $0 < abs(x-1) < delta$ we have $abs((2x-4)-(-2)) < epsilon$ .

So, by the definition of limit, $lim_(xrarr3)(2x-4) = -2$ .

How do you use the epsilon delta definition to prove that the limit of 2x-4=62x-4=6 as x->1x->1?