How do you find the Limit of #x+3# as #x->2# and then use the epsilon delta definition to prove that the limit is L?

To evaluate, we can simply take the limit by substituting #2# in for #x#, and we then have #(2+3)=5#.

Now, prove #lim_(x->2)x+3=5#.

The epsilon delta proof for limits is easier understood when one is familiar with the definitions of the terms involved. Most useful will be the definition of the limit of a function.

#lim_(x->c)f(x)=L#

Definition:

• Let #f: D->RR#
• Let #c# be an accumulation point of #D#
• We say that the limit of #f(x)# at #c# is the real number #L# provided that:

for every #epsilon>0#, there exists #delta>0# such that #abs(f(x)-L)< epsilon# whenever #0< abs(x-c)< delta#.

In our case:

• #f(x)=x+3#
• #c=2#
• #L=5#

We then need to find an expression for #delta# so that the definition, and consequently the proof, holds. In other words, we need:

#abs(x+3-5)< epsilon" "# when #" "0< abs(x-2)< delta#

So:

#abs(x-2)< epsilon#

Since the concept of the limit only applies when #x# is close to #a#, we will need to restrict #x# so that it is at most #1# away from #a#, or: #abs(x-a) <1#. For us, that is #abs(x-2)< 1#.

In this case, we have #x-2# for both cases, so we need not evaluate both.

Then we know that we must have:

#abs(x-2)< delta=epsilon#

#=>delta=epsilon#

Note that all of the above steps come from the definition of the limit of a function as provided above.

Since we have #delta=epsilon# and both cases were the same, the proof will be pretty short.

Proof. Let #epsilon>0# be arbitrary and let #delta=epsilon#. If #-1 < abs(x-2)< delta#, then #abs((x+3)-2)=abs(x-2) <1#; hence:

#abs(x-2) <1#

#<= delta=epsilon" " square#