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

Question

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

Christopher Allers · Accepted Answer

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#

Isaiah Allen · Answer

undefined To prove that lim_(x->1)(x+2)= 3 using the epsilon delta definition of limit, we need to show that for any given epsilon > 0, there exists a delta > 0 such that if 0 < |x - 1| < delta, then |(x+2) - 3| < epsilon. Let's proceed with the proof: Given epsilon > 0, we need to find a suitable delta > 0. Now, |(x+2) - 3| = |x - 1|. We want to ensure that |x - 1| < delta implies |(x+2) - 3| < epsilon. Let's choose delta = epsilon. If 0 < |x - 1| < delta = epsilon, then |(x+2) - 3| = |x - 1| < epsilon. Hence, we have shown that for any epsilon > 0, there exists a delta > 0 (specifically, delta = epsilon) such that if 0 < |x - 1| < delta, then |(x+2) - 3| < epsilon. Therefore, by the epsilon delta definition of limit, we can conclude that lim_(x->1)(x+2) = 3.

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

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