Let #epsilon > 0# be arbitrary. We wish to show that there exists a #delta > 0# such that if #0 < |x-2| < delta# then #|sqrt(x+7)-3| < epsilon#.
Suppose #delta = min{epsilon, 9}#, and that #0 < |x-2| < delta#.
First, note that #-delta < x-2 < delta#
#=> x+7 > 9-delta#
#=> sqrt(x+7) > sqrt(9-delta)#
#=> sqrt(x+7)+3 > sqrt(9-delta)+3 >= 3 > 1#
Now, proceeding:
#|sqrt(x+7)-3| = |(sqrt(x+7)-3)*(sqrt(x+7)+3)/(sqrt(x+7)+3)|#
#=|(x+7-9)/(sqrt(x+7)+3)|#
#=|x-2|/|sqrt(x+7)+3|#
#<|x-2|/1" "# (as #sqrt(x+7)+3 > 1#)
#= |x-2|#
#< epsilon" "# (as #delta <= epsilon# by our choice of #delta#)
Thus our choice of #delta# satisfies the desired conditions. ∎
