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

Sep 24, 2014

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
$0 < | x - 1 | < \delta R i g h t a r r o w | \left(x + 2\right) - 3 | < \epsilon$.

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

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