How do you use the epsilon delta definition of limit to prove that $lim_(x->1)(x+2)= 3$ ?

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Answer 1 · 2014-09-24T19:54:06

Before writing a proof, I would do some scratch work in order to find the expression for $delta$ in terms of $epsilon$ .

According to the epsilon delta definition, we want to say:

For all $epsilon > 0$ , there exists $delta > 0$ such that
$0<|x-1|< delta Rightarrow |(x+2)-3| < epsilon$ .

Start with the conclusion.

$|(x+2)-3| < epsilon Leftrightarrow |x-1| < epsilon$

So, it seems that we can set $delta =epsilon$ .

(Note: The above observation is just for finding the expression for $delta$ , so you do not have to include it as a part of the proof.)

Here is the actual proof:

Proof

For all $epsilon > 0$ , there exists $delta=epsilon > 0$ such that
$0<|x-1| < delta Rightarrow |x-1|< epsilon Rightarrow |(x+2)-3| < epsilon$

How do you use the epsilon delta definition of limit to prove that lim_(x->1)(x+2)= 3lim_(x->1)(x+2)= 3 ?