How do you use the epsilon delta definition to prove that the limit of #(1+x)/(4+x)=1/2# as #x->2#?

1 Answer
Jim H
Feb 23, 2017

See below.

Explanation:

Given a positive #epsilon#, let #delta = min{1, 14epsilon}#. Note that #delta# is positive.

For any #x# with #0 < abs(x-2) < delta#, we have

#abs(x-2) < 1#, which is equivalent to #-1 < x-2 < 1#, and this entails that #-5 < x+4 < 7#, so we will have #-10 < 2(x+4) < 14# and, finally we are assured that #abs(2(x+4)) < 14#

So for any #x# with #0 < abs(x-2) < delta#, we have

#abs((1+x)/(4+x) - 1/2) = abs((2+2x-4-x)/(2(x+4)))#

# = abs((x-2)/(2(x+4))#

# < (abs(x-2))/14#

# < delta/14#

# < (14epsilon)/14#

# = epsilon#

We have shown that for any #epsilon > 0# there is a #delta > 0# such that

for any #x# with #0 < abs(x-2) < delta#

#abs((1+x)/(4+x) - 1/2)< epsilon#

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